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<bits/stdc++.h>
using namespace std;
int main(){
  long N,K;cin>>N>>K;string S;cin>>S;
  
  vector<int> cumsum(1,1);
  for(int i=1;i<N;i++){
    if(S[i]!=S[i-1])cumsum.push_back(cumsum[cumsum.size()-1]);
    cumsum[cumsum.size()-1]++;
  }
  if(2*K-(S[0]=='0')>=cumsum.size()){cout<<N;return 0;}
  int s=cumsum[2*K-(S[0]=='0')];
  
  int ans=s;
  int l=(S[0]=='1'),r=2*K-(S[0]=='0')+2;
  while(r<cumsum.size()){
    ans=max(ans,cumsum[r]-cumsum[l]);
    l+=2;r+=2;
  }
  if(S[N-1]=='0'){
    ans=max(ans,cumsum[cumsum.size()-1]-cumsum[cumsum.size()-1-2*K]);
  }
  cout<<ans<<endl;
}