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;
  string s;
  cin>>n>>k>>s;
  s+='#';
  vector<int> v;
  int c=0;
  char h='1';
  for(int i=0;i<n+1;i++){
    if(h!=s.at(i)){
      v.push_back(c);
      h=s.at(i);
      c=0;
    }
    c++;
  }
  if(v.size()%2==0)
    v.push_back(0);
  int a=0;
  for(int i=0;i<k*2+1&&i<v.size();i++)
    a+=v.at(i);
  int m=a;
  for(int i=k*2+2;i<v.size();i+=2){
    a+=v.at(i-1)+v.at(i)-v.at(i-k*2-2)-v.at(i-k*2-1);
    m=max(m,a);
  }
  cout<<m<<endl;
}