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: Runtime Error)

#include <stdio.h>
int max(int a,int b) {return a>b?a:b;}
int run(int *u, int l, int k) {
	int s[100001] = {u[0]}, m = u[0], i;
	for(i = 1; i < l; i++) {
		s[i] = s[i - 1] + u[i];
		if(!(i % 2) && i >= k*2) {
			m = max(m, s[i] - (i == k*2 ? 0 : s[i - k*2 - 1]));
		} else if(i == l - 1)
			m = max(m, s[i] - (i + 1 == k*2 ? 0 : s[i - k*2]));
	}
	return m;
}
int main() {
	int n, k, s[100000], u[100001], l = 0, m, c = 1, i;
	char t;
	scanf("%d %d ", &n, &k);
	for(i = 0; i < n; i++) {
		scanf("%c", &t);
		s[i] = t - '0';
		if(!i) {
			m = s[i];
			if(!m)
				u[l++] = 0;
		} else if(s[i] == m)
			c++;
		else {
			u[l++] = c;
			m = !m;
			c = 1;
		}
	}
	u[l++] = c;
	
	printf("%d", run(u, l, k));
	return 0;
}