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<algorithm>
#include<vector>
#include<map>
#include<string>

using namespace std;
int main(){
int N,K;
cin >>N>>K;

string s;
cin >> s;

int left[300010];
left[1]=1;
char c=s[0];
int l=1;
 for(int i=1;i<s.size();i++){
    if(s[i]==c){
        continue;
    }
    else{
        c=s[i];
        left[++l]=i+1;       
    }
}
for(int i=l+1;i<=300000;i++){
    left[i]=N+1;
}
 int x;
 int ans=0;
for(int k=1;k<=l;k++){
    if(s[left[k]-1]=='0'){
        x=left[k+2*K]-left[k];
        ans=max(x,ans);
    }
    else{
        x=left[k+2*K+1]-left[k];
        ans=max(x,ans);

    }
}
cout << ans <<endl;

}