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>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)(n);i++)

int main(){
  int N, K;
  string S;
  cin >> N >> K >> S;

  vector<int> start;
  REP(i, N) {
    if((i == 0) || (S[i] != S[i - 1])) {
      start.push_back(i);
    }
  }
  start.push_back(N);
  int M = start.size();

  int ret = 0;
  REP(i, M - 1) {
    int tmp;
    if(S[start[i]] == '0') {
      tmp = start[min(M - 1, i + 2 * K)] - start[i];
    }else{
      tmp = start[min(M - 1, i + 2 * K + 1)] - start[i];
    }
    ret = max(ret, tmp);
  }

  cout << ret << endl;
  return 0;
}
