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 pii pair<int,int>
#define F first
#define S second
#define mp make_pair
#define pb push_back
using namespace std;
const int N=1e5+5;
vector<int>v;
int n,k,a[N],ji=0,last,ans=-1,dp[N];
string s;
int main()
{
	cin>>n>>k;
	cin>>s;
	if(s[0]=='0')v.pb(0);
	last=s[0]-'0';
	for(int i=0;i<s.size();i++)
	{
		if(s[i]-'0'!=last)
		{
			v.pb(ji);
			last=s[i]-'0';
			ji=1;
		}
		else
			ji++;
	}
	v.pb(ji);
	for(int i=1;i<=v.size();i++)dp[i]=dp[i-1]+v[i-1];
	for(int i=0;i<v.size();i+=2)
	{
		int pnt=dp[min(i+2*k+1,(int)v.size())]-dp[i];
		ans=max(ans,pnt);
	}
	cout<<ans<<endl;
}