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;
int main()
{
string s;
int i,j,k,n,m,l,flag,w,max;
vector<int> left(200000),right(20000000);
cin>>n >> k;
cin >> s ;
flag=1;
w=1;
if(s.at(0)=='1')
{
left.at(1)=0;
flag++;
}
for(i=0;i<n-1;i++)
{
if(s.at(i)=='0'&&s.at(i+1)=='1')
{
left.at(flag)=i+1;
flag++;
}
if(s.at(i)=='1'&&s.at(i+1)=='0')
{
right.at(w)=i;
w++;
}
}
if(s.at(n-1)=='1')
{
right.at(w)=n-1;
w++;
}
left.at(0)=0;
right.at(w)=n-1;
//cout << flag-1 << w-1 <<endl;
m=w-1;
max=0;
if(w>=k)
{
for(i=0;i<=w-k;i++)
{
l=right.at(i+k)-left.at(i)+1;
if(l>=max)
{
max=l;
}
}
}
else
{
max=n;
}
cout << max <<endl;
}