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 < (int)(n); ++i)
#define FOR(i, k, n) for (int i = (k); i < (int)(n); ++i)
typedef long long int ll;

using namespace std;

int main() {
  int N, K;
  string S;
  cin >> N >> K >> S;
  vector<int> I({1});
  FOR(i, 1, N) {
    if (S[i] != S[i - 1]) I.push_back(i + 1);
  }

  int ans = 0;
  int r = I.size();
  REP(k, r) {
    int cnt;
    if (S[I[k] - 1] == '0') {
      cnt = (k + 2 * K >= r ? N + 1 : I[k + 2 * K]) - I[k];
    } else {
      cnt = (k + 2 * K + 1 >= r ? N + 1 : I[k + 2 * K + 1]) - I[k];
    }
    ans = max(ans, cnt);
  }
  cout << ans << endl;
}