Good Substrings CodeForces - 271D

You've got string s, consisting of small English letters. Some of the English letters are good, the rest are bad.

A substring s[l...r] (1 ≤ l ≤ r ≤ |s|) of string s  =  s₁s₂...s_|s| (where|s| is the length of string s) is string  s_ls_l + 1...s_r.

The substring s[l...r] is good, if among the letters  s_l, s_l + 1, ..., s_rthere are at most k bad ones (look at the sample's explanation to understand it more clear).

Your task is to find the number of distinct good substrings of the given string s. Two substrings s[x...y] and s[p...q] are considered distinct if their content is different, i.e. s[x...y] ≠ s[p...q].

Input

The first line of the input is the non-empty string s, consisting of small English letters, the string's length is at most 1500characters.

The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the i-th character of this string equals "1", then the i-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.

The third line of the input consists a single integer k (0 ≤ k ≤ |s|) — the maximum acceptable number of bad characters in a good substring.

Output

Print a single integer — the number of distinct good substrings of string s.

Examples

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Input

ababab
01000000000000000000000000
1

Output

5

Input

acbacbacaa
00000000000000000000000000
2

Output

8

Note

In the first example there are following good substrings: "a", "ab", "b", "ba", "bab".

In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".

题意：已知26个字母的好坏，

要求从给出的一个字符串中找出满足条件的子串数。

【子串需满足：其存在的坏字母数最多只有k个】

　　

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 1e9 + 7;
 5 const int maxn = 1e4 + 10;
 6 string  str1, str2;
 7 int k, a[maxn];
 8 LL seed[3]={13131, 10007, 11137};
 9 int main() {
10     cin >> str1 >> str2 >> k;
11     a[0] = (str2[str1[0] - 'a'] == '0');
12     for (int i = 1 ; i < str1.size() ; i++)
13         a[i] = a[i - 1] + (str2[str1[i] - 'a'] == '0');
14     set<LL>st;
15     for (int i = 0 ; i < str1.size() ; i++) {
16         LL HASH = str1[i] - 'a' + 1;
17         for (int j = i ; j < str1.size() ; j++) {
18             if (i == j && (str2[str1[j] - 'a'] == '0') <= k ) st.insert(str1[j] - 'a');
19             else {
20                 if (a[j] - a[i] + (str2[str1[i] - 'a'] == '0') > k) break;
21                 HASH += (HASH * seed[str1[j] % 3] + str1[j] - 'a') % mod;
22                 st.insert(HASH);
23             }
24         }
25     }
26     printf("%d\n", st.size());
27     return 0;
28 }