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 <utility>
#include <string>
#include <vector>

int main(int, char**) {
  int n, K;
  std::string s;
  std::cin >> n >> K >> s;

  std::vector<int> is;
  char pre{};
  for(int i{}; i < n; ++i) {
    if(pre != s[i]) is.push_back(i);

    pre = s[i];
  }

  int ans{-1};
  size_t r{is.size()};
  for(int k{}; k < r; ++k) {
    int x;
    if(s[is[k]] == '0') {
      x = ((k + 2 * K >= r) ? n : is[k + 2 * K]) - is[k];
    } else {
      x = ((k + 2 * K + 1 >= r) ? n : is[k + 2 * K + 1]) - is[k];
    }

    ans = std::max(x, ans);
  }

  std::cout << ans << std::endl;

  return 0;
}
