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 <utility>
using namespace std;

int main(){
  int N, K;
  string S;
  cin >> N >> K >> S;
  vector<pair<char,int>> V;
  for(int i = 0; i < N; ++i){
    if(V.empty() or V.back().first != S[i]) V.emplace_back(S[i],1);
    else ++V.back().second;
  }
  int M = V.size();
  vector<int> C(M+1,0);
  for(int i = 0; i < M; ++i){
    C[i+1] = C[i]+V[i].second;
  }
  int L = 2*K, ans = 0;
  for(int i = 0; i < M; ++i){
    if(V[i].first == '1'){
      ans = max(ans,C[min(M,i+L+1)]-C[i]);
    }else{
      ans = max(ans,C[min(M,i+L)]-C[i]);
    }
  }
  cout << ans << endl;
}
