Codeforces Round #478 (Div. 2) ABC

A. Aramic script

In Aramic language words can only represent objects.

Words in Aramic have special properties:

• A word is a root if it does not contain the same letter more than once.
• A root and all its permutations represent the same object.
• The root $\mathrm{xx of\; a\; word yy is\; the\; word\; that\; contains\; all\; letters\; that\; appear\; in yy in\; a\; way\; that\; each\; letter\; appears\; once.\; For\; example,\; the rootof\; "aaaa",\; "aa",\; "aaa"\; is\; "a",\; the root of\; "aabb",\; "bab",\; "baabb",\; "ab"\; is\; "ab".}$
• Any word in Aramic represents the same object as its root.

You have an ancient script in Aramic. What is the number of different objects mentioned in the script?

Input

The first line contains one integer $\mathrm{nn (1\le n\le 1031\le n\le 103) —\; the\; number\; of\; words\; in\; the\; script.}$

The second line contains $\mathrm{nn words s1,s2,\dots ,sns1,s2,\dots ,sn —\; the\; script\; itself.\; The\; length\; of\; each\; string\; does\; not\; exceed 103103.}$

It is guaranteed that all characters of the strings are small latin letters.

Output

Output one integer — the number of different objects mentioned in the given ancient Aramic script.

Examples

input

Copy

5
a aa aaa ab abb

output

Copy

2

input

Copy

3
amer arem mrea

output

Copy

1

Note

In the first test, there are two objects mentioned. The roots that represent them are "a","ab".

In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

 查找有多少个root，英语渣渣的我看了好久才看到，ab,ba 是一个单词，它的原根是ab，求原根的话是将一个单词重复的字符去掉，比如aaabbb的原根是ab，因为只出现a字符和b字符。懂意思就好做了。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 1010;
 5 set<int> st;
 6 string s;
 7 int n;
 8 int main() {
 9     cin >> n;
10     int ans = 0;
11     for(int i = 0; i < n; i ++) {
12         cin >> s;
13         int cnt = 0;
14         for(int j = 0; j < s.length(); j ++) {
15             cnt |= (1<<(s[j]-'a'));
16         }
17         st.insert(cnt);
18     }
19     cout << st.size() << endl;
20     return 0;
21 }

B. Mancala

Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.

Initially, each hole has ${\mathrm{aiai stones.\; When\; a\; player\; makes\; a\; move,\; he\; chooses\; a\; hole\; which\; contains\; a positive number\; of\; stones.\; He\; takes\; all\; the\; stones\; inside\; it\; and\; then\; redistributes\; these\; stones\; one\; by\; one\; in\; the\; next\; holes\; in\; a\; counter-clockwise\; direction.}}^{}$

Note that the counter-clockwise order means if the player takes the stones from hole $\mathrm{ii,\; he\; will\; put\; one\; stone\; in\; the (i+1)(i+1)-th\; hole,\; then\; in\; the (i+2)(i+2)-th,\; etc.\; If\; he\; puts\; a\; stone\; in\; the 1414-th\; hole,\; the\; next\; one\; will\; be\; put\; in\; the\; first\; hole.}$

After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.

Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.

Input

The only line contains 14 integers ${\mathrm{a1,a2,\dots ,a14a1,a2,\dots ,a14 (0\le ai\le 1090\le ai\le 109) —\; the\; number\; of\; stones\; in\; each\; hole.}}^{}$

It is guaranteed that for any $\mathrm{ii (1\le i\le 141\le i\le 14) aiai is\; either\; zero\; or\; odd,\; and\; there\; is\; at\; least\; one\; stone\; in\; the\; board.}$

Output

Output one integer, the maximum possible score after one move.

Examples

input

Copy

0 1 1 0 0 0 0 0 0 7 0 0 0 0

output

Copy

4

input

Copy

5 1 1 1 1 0 0 0 0 0 0 0 0 0

output

Copy

8

Note

In the first test case the board after the move from the hole with $\mathrm{77 stones\; will\; look\; like 1\; 2\; 2\; 0\; 0\; 0\; 0\; 0\; 0\; 0\; 1\; 1\; 1\; 1.\; Then\; the\; player\; collects\; the\; even\; numbers\; and\; ends\; up\; with\; a\; score\; equal\; to 44.}$

 模拟，14个洞，从第i个洞取出全部石头，然后在第i+1上放一个，第i+2上放一个，循环直到全部放完，然后把14个洞里偶数的数相加，求最大值。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 110;
 5 ll a[15], b[15];
 6 void init() {
 7     for(int i = 0; i < 14; i ++) b[i] = a[i];
 8 }
 9 int main() {
10     for(int i = 0; i < 14; i ++) {
11         cin >> a[i];
12     }
13     ll MAX = 0;
14     for(int i = 0; i < 14; i ++) {
15         init();
16         ll x = b[i];
17         b[i] = 0;
18         for(int j = 0; j < 14; j ++) {
19             b[j] += x/14;
20         }
21         x = x%14;
22         for(int j = 0; j < x; j ++) {
23             b[(i+j+1)%14] ++;
24         }
25         ll cnt = 0;
26         for(int j = 0; j < 14; j ++) {
27             if(b[j]%2==0) cnt += b[j];
28         }
29         MAX = max(MAX, cnt);
30     }
31     // for(int i = 0; i < 14; i ++) printf("%lld ",b[i]);printf("\n");
32     cout << MAX << endl;
33     return 0;
34 }

C. Valhalla Siege

Ivar the Boneless is a great leader. He is trying to capture Kattegat from Lagertha. The war has begun and wave after wave Ivar's warriors are falling in battle.

Ivar has $\mathrm{nn warriors,\; he\; places\; them\; on\; a\; straight\; line\; in\; front\; of\; the\; main\; gate,\; in\; a\; way\; that\; the ii-th\; warrior\; stands\; right\; after (i-1)(i-1)-th\; warrior.\; The\; first\; warrior\; leads\; the\; attack.}$

Each attacker can take up to ${\mathrm{aiai arrows\; before\; he\; falls\; to\; the\; ground,\; where aiai is\; the ii-th\; warrior\text{'}s\; strength.}}^{}$

Lagertha orders her warriors to shoot ${\mathrm{kiki arrows\; during\; the ii-th\; minute,\; the\; arrows\; one\; by\; one\; hit\; the\; first\; still\; standing\; warrior.\; After\; all\; Ivar\text{'}s\; warriors\; fall\; and\; all\; the\; currently\; flying\; arrows\; fly\; by,\; Thor\; smashes\; his\; hammer\; and\; all\; Ivar\text{'}s\; warriors\; get\; their\; previous\; strengths\; back\; and\; stand\; up\; to\; fight\; again.\; In\; other\; words,\; if\; all\; warriors\; die\; in\; minute tt,\; they\; will\; all\; be\; standing\; to\; fight\; at\; the\; end\; of\; minute tt.}}^{}$

The battle will last for $\mathrm{qq minutes,\; after\; each\; minute\; you\; should\; tell\; Ivar\; what\; is\; the\; number\; of\; his\; standing\; warriors.}$

Input

The first line contains two integers $\mathrm{nn and qq (1\le n,q\le 2000001\le n,q\le 200000) —\; the\; number\; of\; warriors\; and\; the\; number\; of\; minutes\; in\; the\; battle.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109)\; that\; represent\; the\; warriors\text{'}\; strengths.}$

The third line contains $\mathrm{qq integers k1,k2,\dots ,kqk1,k2,\dots ,kq (1\le ki\le 10141\le ki\le 1014),\; the ii-th\; of\; them\; represents\; Lagertha\text{'}s\; order\; at\; the ii-th\; minute: kiki arrows\; will\; attack\; the\; warriors.}$

Output

Output $\mathrm{qq lines,\; the ii-th\; of\; them\; is\; the\; number\; of\; standing\; warriors\; after\; the ii-th\; minute.}$

Examples

input

Copy

5 5
1 2 1 2 1
3 10 1 1 1

output

Copy

3
5
4
4
3

input

Copy

4 4
1 2 3 4
9 1 10 6

output

Copy

1
4
4
1

Note

In the first example:

• after the 1-st minute, the 1-st and 2-nd warriors die.
• after the 2-nd minute all warriors die (and all arrows left over are wasted), then they will be revived thus answer is 5 — all warriors are alive.
• after the 3-rd minute, the 1-st warrior dies.
• after the 4-th minute, the 2-nd warrior takes a hit and his strength decreases by 1.
• after the 5-th minute, the 2-nd warrior dies.

二分查找。

有n个战士，每个战士都有个值ai，表示需要射ai只箭才死，从第一个战士开始射，全部死后就会全部复活。

每射k只箭，求还活着的战士。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2e5+10;
 5 ll a[N], n, q, sum[N];
 6 int main() {
 7     cin >> n >> q >> a[0];
 8     for(int i = 1; i < n; i ++) {
 9         cin >> a[i];
10         a[i] += a[i-1];
11     }
12     ll id = 0, ans = 0;
13     while(q--) {
14         ll k;
15         cin >> k;
16         ans += k;
17         id = upper_bound(a,a+n,ans)-a;
18         if(id == n) id = 0, ans = 0;
19         cout << n - id << endl;
20     }
21     return 0;
22 }