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>
#define MB template <class T>
#define Fur(i,x,y) for(int i=x;i<=y;i++)
#define fcin ios::sync_with_stdio(false)
MB T MAX(T x,T y){return x>y?x:y;}
using namespace std;
#define N 400000
char ch[N];
int n,k,a[N],d=1,ans=0,t;
int main(){
    fcin;
    cin>>n>>k>>ch;
    for(int i=0;i<n;){
        while(ch[i]==49)a[d]++,i++;d++;
        while(ch[i]==48)a[d]++,i++;d++;
    }
    Fur(i,1,d)a[i+1]+=a[i];
    t=k*2+1;
    Fur(i,1,d)ans=MAX(ans,a[i]-a[MAX(i-t,0)]),i++;
    cout<<ans<<endl;
}