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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int main(){

    int N, K, i, j, maxans = 0;
    scanf("%d %d", &N, &K);

    char S_[N];
    scanf("%s", &S_);
    int S[N];
    for (i=0; i<N; i++){
        S[i] = S_[i] - '0';
    }

    int P1[N], P2[N], zf = 0, cnt = 0;
    int *num = (int*)calloc(N, sizeof(int));
    for (i=0; i<N; i++){
        if ((S[i] == 0) && (zf == 0)){
            P1[cnt] = i;
            zf = 1;
        }
        if ((S[i+1] == 1 || i+2 > N) && (zf == 1)){
            P2[cnt] = i;
            num[cnt] =  P2[cnt] + 1 - P1[cnt];
            cnt++;
            zf = 0;
        }
    }

    int P1_[N], P2_[N], cnt_ = 0;
    int *num_ = (int*)calloc(N, sizeof(int));
    zf=0;
    for (i=0; i<N; i++){
        if ((S[i] == 1) && (zf == 0)){
            P1_[cnt_] = i;
            zf = 1;
        }
        if ((S[i+1] == 0 || i+2 > N) && (zf == 1)){
            P2_[cnt_] = i;
            num_[cnt_] =  P2_[cnt_] + 1 - P1_[cnt_];
            if (num_[cnt_] > maxans) maxans = num_[cnt_];
            cnt_++;
            zf = 0;
        }
    }


    int ans;
    for (i=0; i<cnt; i++){
        ans = 0;
        for (j=0; j<K; j++){
            if (j>=cnt) break;
            ans += (num_[i+j] + num[i+j]);
        }
        if (S[0] == 1) ans += num_[i+K];
        if (S[0] == 0 && i-1 > 0) ans += num_[i-1];
        if (maxans < ans) maxans = ans;
    }


    printf("%d", maxans);

    return 0;
}