Substrings Sort string 基本操作

You are given $\mathrm{nn strings.\; Each\; string\; consists\; of\; lowercase\; English\; letters.\; Rearrange\; (reorder)\; the\; given\; strings\; in\; such\; a\; way\; that\; for\; every\; string,\; all\; strings\; that\; are\; placed\; before\; it\; are\; its substrings.}$

String $\mathrm{aa is\; a substring of\; string bb if\; it\; is\; possible\; to\; choose\; several consecutive letters\; in bb in\; such\; a\; way\; that\; they\; form aa.\; For\; example,\; string\; "for"\; is\; contained\; as\; a substring in\; strings\; "codeforces",\; "for"\; and\; "therefore",\; but\; is\; not\; contained\; as\; a substring in\; strings\; "four",\; "fofo"\; and\; "rof".}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 1001\le n\le 100)\; —\; the\; number\; of\; strings.}$

The next $\mathrm{nn lines\; contain\; the\; given\; strings.\; The\; number\; of\; letters\; in\; each\; string\; is\; from 11 to 100100,\; inclusive.\; Each\; string\; consists\; of\; lowercase\; English\; letters.}$

Some strings might be equal.

Output

If it is impossible to reorder $\mathrm{nn given\; strings\; in\; required\; order,\; print\; "NO"\; (without\; quotes).}$

Otherwise print "YES" (without quotes) and $\mathrm{nn given\; strings\; in\; required\; order.}$

Examples

input

Copy

5
a
aba
abacaba
ba
aba

output

Copy

YES
a
ba
aba
aba
abacaba

input

Copy

5
a
abacaba
ba
aba
abab

output

Copy

NO

input

Copy

3
qwerty
qwerty
qwerty

output

Copy

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 typedef long long LL;
 5 int n;
 6 string a[101];
 7 int cmp(string a, string b) {
 8     return a.size() < b.size();
 9 }
10 int main() {
11     scanf("%d", &n);
12     for (int i = 0 ; i < n ; i++) cin >> a[i];
13     sort(a, a + n, cmp);
14     int num=0;
15     for (int i = 1 ; i < n ; i++) {
16         if (a[i].find(a[i-1]) != string::npos) num++;
17     }
18     if (num!=n-1) printf("NO\n");
19     else {
20         printf("YES\n");
21         for (int i = 0 ; i < n ; i++ )
22             cout << a[i] << endl;
23     }
24     return 0;
25 }

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