Another Problem on Strings

Description

A string is binary, if it consists only of characters "0" and "1".

String v is a substring of stringw if it has a non-zero length and can be read starting from some position in stringw. For example, string "010" has six substrings: "0", "1", "0", "01", "10", "010". Two substrings are considered different if their positions of occurrence are different. So, if some string occurs multiple times, we should consider it the number of times it occurs.

You are given a binary string s. Your task is to find the number of its substrings, containing exactlyk characters "1".

Input

The first line contains the single integer k (0 ≤ k ≤ 106). The second line contains a non-empty binary strings. The length of s does not exceed 106 characters.

Output

Print the single number — the number of substrings of the given string, containing exactlyk characters "1".

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use thecin, cout streams or the %I64d specifier.

Sample Input

Input
1
1010

Output

6
Input
2
01010
Output
4

Input

100
01010

Output
0
题意大概就是 有许多的01的串，然后找出包含输入的k个1的子串的总数
这道题 用dp[j]表示，以第目前（循环到达的位置i，不改变j的大小的情况下后面的0的个数

int k, cnt;
    while(scanf("%d %s", &k, s) != EOF){
        cnt = 0;
        CLR(dp);
        dp[0] = 1;//加1是因为，就是首位没有0.也有一种子串
        ll ans = 0;
        int len = strlen(s);
        for(int i = 0; i < len; i++){
            if(s[i] == '1') cnt++;//首尾就算非0也加1
            if(cnt >= k) ans += dp[cnt - k];
            dp[cnt]++;//首位之后0的个数 + 1,