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 <iostream>
#include <vector>
using namespace std;
int main(void){
    int n, k, a=0;
    string s;
    char c='a';
    cin>>n>>k>>s;
    int w=0;
    vector<int> v;
    v.push_back(0);
    for(int i=0;i<n;i++){
        if(s[i]!=c){
            if(w>0)v.push_back(w);
            c=s[i];
        }
        w++;
        if(i==n-1)v.push_back(w);
    }
    int u=v.size(), l=0, r=k*2+1;
    if(s[0]=='0'){
        a=max(a, v[min(k*2, u-1)]-v[0]);
        l=1;
        r=2*k+2;
    }
    while(l<u-1){
        r=min(r, u-1);
        a=max(a, v[r]-v[l]);
        l+=2;
        r+=2;
    }
    cout<<a<<endl;
}
