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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;
typedef long long ll;

int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int ans = 0;
  int right = 0;
  int cn = 0;
  if (s[0] == '0') cn++;
  for (int left = 0; left < n; ++left){
    while (cn <= k && right < n){
      if(right + 1 < n && s[right] == '1' && s[right +1] == '0')cn++;
      ++right;
    }
    /* break した状態で right は条件を満たす最大なので、何かする */
    ans = max(ans,right - left);

    if (left + 1 < n && s[left] == '0' && s[left + 1] == '1') cn--;
  }
  cout << ans << endl;
}