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.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
#include<cstdio>
#define N 100005
using namespace std;
int max(int a,int b){
	return a>b?a:b;
}
char s[N];
int cnt,n;
int getnext(char qw){
	for(int i=0;;++i){
		if(s[cnt+i]!=qw){
			cnt+=i;
			return i;
		}
		if(cnt+i==n){
			cnt+=i;
			return i;
		}
	}
}
int solve[N];
int main(){
	int k;
	scanf("%d%d%s",&n,&k,s);
	int count=0;
	for(;cnt!=n;++count){
		solve[count]=getnext('1');
		++count;
		solve[count]=getnext('0');
	}
	int ans=0,tmp=solve[0];
	for(int i=1;i<count;i+=2){
		tmp+=solve[i]+solve[i+1];
		if((i>>1)>=k){
			tmp-=solve[i-(k<<1)-1]+solve[i-(k<<1)];
		}
		ans=max(ans,tmp);
	}
	printf("%d",ans);
}