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

int main(){
    int n,k;string ss;cin>>n>>k>>ss;
    vector<ll> a,s(1,0);

    int count=1;
    for(int i=1;i<n;i++){
        if(ss.at(i-1)!=ss.at(i)){
            a.push_back(count);
            count=1;
        }else count++;
    }
    a.push_back(count);

    int na=a.size();
    REP(i,na) s.push_back(s.at(i)+a.at(i));

    ll ans=0;
    for(int i=ss.at(0)-'0'-1;i<na;i+=2){
        int l=max(0,i),r=min(i+2*k+1,na);
        ans=max(ans,s.at(r)-s.at(l));
    }
    cout<<ans;
}