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 <vector>
#include <algorithm>
#include <cstring>
using namespace std;

int main() {
  int N,K,H;
  cin>>N>>K;
  string S;
  cin>>S;
  H = N+2*K+2;
  vector <int> v(H);
  for (int i = 0; i < H; i++) v[i] = N;
  v[0] = 0;
  int j = 1;
  for (int i = 1; i < N; i++) {
    if (S[i] != S[i-1])  {
      v[j] = i;
      j++;
    }
  }
  int ans;
  ans = 0;
  for (int i = 0; i < j; i++) {
    if (S[v[i]]=='0') ans = max(v[i+2*K]-v[i],ans);
    else if (S[v[i]]=='1') ans = max(v[i+2*K+1]-v[i],ans);
  }
  cout<<ans<<endl;
  return 0;
}