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<iostream>
#include<string>

std::string s;

int n,k,d[100002],res,count,sum;

int main(){

    std::cin>>n>>k>>s;

    for(int i=0;i<n;i++){
        d[i]=1;
    }

    if(s[0]=='0')
    d[0]=0,count++;

    for(int i=1;i<n;i++){
        if(s[i]==s[i-1])
        d[count]++;
        else
        count++;
    }

    if(count%2==1)
    d[count+1]=0,count++;

    if(k*2>count)
    k=count/2;

    for(int i=0;i<=(count-k*2)/2;i++){
        sum=0;
        for(int j=i*2;j<i*2+k*2+1;j++){
            sum+=d[j];
        }
        if(res<sum)
        res=sum;
    }

    std::cout<<res<<std::endl;

    return 0;

}