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(){
  int N, K, l, m = 0;
  string S, s, a = "10", b = "01";
  vector<int> A, B(1);
  bool flag = true;
  cin >> N >> K;
  cin >> S;
  
  if (S.at(0) == '0'){
    A.push_back(0);
  }
  
  for (int i = 0; i < N-1; i++){
    s = S.substr(i,2);
    if (s == b){
      B.push_back(i+1);
    }
    else if (s == a){
      A.push_back(i+1);
    }
  }
  
  A.push_back(N);
  l = A.size();
  
  for (int i = 0; i < l-K; i++){
    m = max(A.at(min(i+K, l-1)) - B.at(i), m);
    flag = false;
  }
  if (flag){
    cout << N << endl;
  }
  else{
    cout << m << endl;
  }
}