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>
#include <stdlib.h>
int main(void){
	int n,k,i;
	scanf("%d%d",&n,&k);
	char s[100002];
	int a[100002];
	int b[100002];
	int cnt;
	scanf("%s",s);
	//連続する1の数を正、0の数を負として配列aに保存する
	i=0;cnt=1;
	a[0]=0;
	while(i<n){
		while(s[i]=='0'){
			a[cnt]--;
			i++;
			if(s[i]=='1'){
				cnt++;
			}
		}
		while(s[i]=='1'){
			a[cnt]++;
			i++;
			if(s[i]=='0'){
				cnt++;
			}
		}
	}
	cnt++;
	b[0]=abs(a[0]);
	for(i=1;i<cnt;i++){
		b[i]+= b[i-1]+abs(a[i]);
	}
	//cnt = 3;
	int ans=0;
	int temp;
	int min ( int x ,int y){
		int min = x;
		if(min>y)min=y;
		return y;
	}
	for(i=1;i<=cnt;i++){
		//左端がiで右端を求めていく。
		int r;
		if(a[i]>0){
			r= i + 2*k;
		}else{
			r = i + 2*k - 1;
		}
		temp = b[r]-b[i-1];
		if(temp>ans)ans=temp;
	}
	printf("%d",ans);
	return 0;
}