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>
#include<algorithm>
char s[100001];
int cnt[100001]={};
int main(){
    int n,k;
    scanf("%d%d%s",&n,&k,s);
    int i=0,cnti=1;
    while(true){
        while(i<n&&s[i]=='0'){
            cnt[cnti]++;
            i++;
        }
        if(i==n)break;
        cnti++;
        while(i<n&&s[i]=='1'){
            cnt[cnti]++;
            i++;
        }
        if(i==n)break;
        cnti++;
    }
    for(i=1;i<=cnti;i++)cnt[i]+=cnt[i-1];
    int ans=cnt[std::min(2*k,cnti)];
    for(i=2;i<=cnti;i+=2){
        ans=std::max(ans,cnt[std::min(i+2*k,cnti)]-cnt[i-1]);
    }
    printf("%d\n",ans);
    return 0;
}