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: Time Limit Exceeded)

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

int main()
{
    int N;
    int K;
    int list[100000];
    int name[100000];
    int s;
    int i;
    int type;
    int index;
    int index2;
    int count;
    int zero_count;
    int max;
    int j;

    scanf("%d %d%*c", &N, &K);

    index = 0;
    s = getchar() - '0';
    type = s;
    if(type == 0) {
        zero_count = 1;
    } else {
        zero_count = 0;
    }
    count = 1;
    for(i = 1; i < N; i++) {
        s = getchar() - '0';
        if(s == type) {
            count += 1;
        } else {
            list[index] = count;
            name[index] = type;
            type = 1 - type;
            if(type == 0) {
                zero_count += 1;
            }
            index += 1;
            count = 1;
        }
    }
    list[index] = count;
    name[index] = type;
    if(type == 0) {
        zero_count += 1;
    }
    index += 1;

    if(zero_count <= K) {
        max = N;
    } else {
        max = 0;
        index2 = index - K;
        for(i = 0; i < index2; i++) {
            zero_count = 0;
            count = 0;
            for(j = i; j < index; j++) {
                count += list[j];
                if(name[j] == 0) {
                    zero_count += 1;
                } else if(zero_count == K) {
                    break;
                }
            }
            if(max < count) {
                max = count;
            }
        }
    }

    printf("%d\n", max);

    return 0;
}