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: Accepted)

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

int main() {
    int n = 0, k = 0;
    scanf("%d %d\n", &n, &k);
    int raw[n+1];
    char s[n+2];
    fgets(s, n+2, stdin);
    int len_raw = 0;
    char c = 0;
    int i = 0;
    while (i < n) {
        raw[len_raw] = 0;
        while ((i < n) && (s[i] == '1')) {
            raw[len_raw]++;
            i++;
        }
        len_raw++;
        raw[len_raw] = 0;
        while ((i < n) && (s[i] == '0')) {
            raw[len_raw]++;
            i++;
        }
        len_raw++;
    }
    raw[len_raw] = 0;
    k = 2 * k - 1;
    int max = 0;
    for (int i=0; i<=k+1; i++) { 
        if (i >= len_raw) break;
        max += raw[i];
    }
    int buf = max;
    for (int i=k; i<len_raw-2; i += 2) {
        buf += raw[i+2] + raw[i+3] - raw[i-k] - raw[i-k+1];
        if (buf > max) max = buf;
    }
    printf("%d\n", max);
    return 0;
}