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;

typedef long long llong;

void solve() {
  llong N, K;
  string s;
  cin >> N >> K >> s;

  vector<llong> ii = {1};

  for(int i = 1; i < N; ++i) {
    if(s[i]!=s[i-1]) { ii.push_back(i+1); }
  }

  llong lmax = 0, r = ii.size(), t;
  for(int k = 0; k < r; ++k) {
    lmax = max(lmax, (s[ii[k]-1]=='0') ?
      (((k+1+2*K > r) ? N+1 : ii[k+2*K]) - ii[k]) :
      (((k+1+2*K+1 > r) ? N+1 : ii[k+2*K+1]) - ii[k]));
  }

  cout << lmax;

}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  solve();
  return 0;
}