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>
using namespace std;

int main(){
  
  int n,k;
  cin>>n>>k;
  string s;
  cin>>s;
  s+='?';
  int a[100000]={0},b[100000]={0};
  int j=0;
  for(int i=0;i<n;i++){
    if(s[i]=='0') b[j]++;
    else a[j]++;
    if(s[i]=='0'&&s[i+1]=='1') j++;
  }
  if(k>j) cout<<n<<endl;
  else{
    int sum=a[k];
    for(int i=0;i<k;i++){
      sum+=a[i];
      sum+=b[i];
    }
    int l=0,ans=sum;
    for(int i=k;i<=j;i++){
      sum-=a[l];
      sum-=b[l];
      l++;
      sum+=a[i+1];
      sum+=b[i];
      ans=max(ans,sum);
    }
    cout<<ans<<endl;
  }
  
}