Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main() {

    int N;
    scanf("%d", &N);

    int K;
    scanf("%d", &K);

    char str[N];
    scanf("%s", str);

    int arr[N];
    int arr_i = 0;
    int sum = 1;
    char pre = str[0];
    for (int i=1; i<N; i++) {
        if (str[i] == pre) {
            sum++;
        } else {
            arr[arr_i] = sum;
            arr_i++;
            sum = 1;
        }
        if (i == N-1) {
            arr[arr_i] = sum;
            arr_i++;
        }
        pre = str[i];
    }

    arr[arr_i] = '\0';

    int max = 1;
    for (int i=0; arr[i]!='\0'; i++) {
        int occur;
        if (i == 0) {
            occur = 2 * K;
        } else {
            occur = 2 * K + 1;
        }
        int cur = 0;
        for (int j=0; arr[i+j]!='\0' && j<occur; j++) {
            cur += arr[i+j];
        }
        if (max < cur) {
            max = cur;
        }
    }

    printf("%d", max);

    return 0;
}