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;
  cin >> N >> K;
  vector<int> S;
  int c = 0;
  char s, bit = '1';
  for (int i = 0; i < N; i++) {
    cin >> s;
    if (s == bit) {
      c++;
    } else {
      S.push_back(c);
      bit = bit == '0' ? '1' : '0';
      c = 1;
    }
  }
  S.push_back(c);

  int SLen = S.size();
  int ans = 0;
  for (int i = 0; i < SLen; i += 2) {
    int count = 0;
    int tempMax = min(SLen, i + K * 2 + 1);
    for (int j = i; j < tempMax; j++) {
      count += S[j];
    }
    ans = max(ans, count);
  }

  cout << ans << endl;

  return 0;
}
