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, i, len=1, ans=1; string S; cin >> N >> K >> S;
vector<int> I, O;
if(S[0]=='0'){ I.push_back(0); }
for(i=1; i<N; i++){ 
  if(S[i-1]==S[i]){ len++; }
  else if(S[i]=='0'){ I.push_back(len); len=1; }
  else if(S[i]=='1'){ O.push_back(len); len=1; }
}
if(S[N-1]=='1'){ I.push_back(len); }
else{ O.push_back(len); I.push_back(0); }
int g=O.size(), h=min(g, K), now=0;
for(i=0; i<h; i++){ now+=(I[i]+O[i]); } now+=I[h]; ans=now;
for(i=h; i<g; i++){ now=now-I[i-h]-O[i-h]+O[i]+I[i+1]; ans=max(ans, now); }  
cout << ans << "\n";
}