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 groups[100002];

int main(void) {
	int n, k;
	scanf("%d%d ", &n, &k);

	int gind = groups[0] = 0;
	for (int i = 0; i < n; i++) {
		int c = getchar() == '1';
		if (c == gind%2) groups[gind]++;
		else groups[++gind] = 1;
	}
	if (gind%2) groups[++gind] = 0;

	int max = 0;
	for (int i = 1; i <= gind; i += 2) {
		if (max < groups[i]) max = groups[i];
	}

	int sum = 0;
	for (int i = 0; i <= gind; i++) {
		if (i > 2*k) sum -= groups[i-2*k-1];
		sum += groups[i];
		if (i%2 != gind%2 && max < sum) max = sum;
	}
	sum = 0;
	for (int i = 0; i <= gind; i++) {
		if (i > 2*k) sum -= groups[gind -(i-2*k-1)];
		sum += groups[gind-i];
		if (i%2 != gind%2 && max < sum) max = sum;
	}

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