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 <vector>
#include <algorithm>
#include <string>
using namespace std;

int main(){
  int N, K, c=0,res=0;
  string S;
  vector<int> vc;
  cin >> N >> K;
  cin >> S;

  vc.push_back(0);
  if(S[0]=='1'){
    vc.push_back(0);
    vc[1]=1;
  }else vc[0]=1;

  for(int i=1;i<N;i++){
    if(S[i]=='0' && S[i-1]=='1' || S[i]=='1' && S[i-1]=='0') vc.push_back(0);
    vc.back()++;
  }

  for(int i=0;i<vc.size();i+=2){
    int sum=0;
    if(i!=0) sum+=vc[i-1];
    for(int j=i;j<=i+(K-1)*2+1 && j<vc.size();j++){
      sum+=vc[j];
    }
    res=max(sum,res);
  }

  cout << res << endl;
  return 0;
}
