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<queue>
using namespace std;
queue<int>q;
int c[100001];
int n,k,cnt,ans;
int main()
{
    c[0]=1;
    cin>>n>>k;
    for(int i=1;i<=n;i++)
    {
        char a;
        cin>>a;
        c[i]=a-48;
        q.push(i);
        if(c[i]==0&&c[i-1]==1)cnt++;
        if(cnt>k)
        {
            cnt--;
            while(c[q.front()]==1)q.pop();
            while(c[q.front()]==0)q.pop();
        }
        //cout<<i-q.front()+1<<endl;
        ans=max(ans,i-q.front()+1);
    }
    cout<<ans;
    return 0;
}