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 s;
    cin>>n>>k>>s;
    int a=1,b=0,c=0;
    rep(i,n){
        if(s[i]=='0'&&a==1){
            if(b==k) break;
            else b++,a=0;
        }
        if(s[i]=='1'&&a==0) a=1;
        c++;
    }
    int d=c,x=0,y=c;
    while(y<n){
        while(s[x]=='1') x++;
        while(s[x]=='0') x++;
        while(s[y]=='0') y++;
        while(s[y]=='1') y++;
        d=max(d,y-x);
    }
    cout<<d;
}