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;
string s;int k;
const int N = 1e5+5;
int n;
vector<int> v;
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>k>>s;
    s="*"+s;
    v.push_back(0);v.push_back(0);
    for(int l=1,r;l<=n;l=r+1){
        r=l;
        if(s[l]=='1')continue;
        while (r+1<=n&&s[r+1]=='0')r++;
        v.push_back(l),v.push_back(r);
    }
    v.push_back(n+1);v.push_back(n+1);
    int ans = 0;
    for(int i=0;i<v.size();i+=2){
        int r = min((int)v.size()-1,i+2*(k+2)-1);
        ans = max(ans,v[r-1]-v[i+1]-1);
    }
    cout<<ans<<endl;
}

