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
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        String S = sc.next();
        
        List<Integer> hs0 = new ArrayList<>();
        List<Integer> hs1 = new ArrayList<>();
        
        if( S.charAt(0) == '0' ) hs1.add(0);
        int count0 = 0;
        int count1 = 0;        
        for(int i = 0; i < N; i++){
            if( S.charAt(i) == '0' ){
                if(count1 > 0) hs1.add(count1);
                count1 = 0;
                count0++;
            }else{
                if(count0 > 0) hs0.add(count0);
                count0 = 0;
                count1++;
            }
            
            if( i == N -1 ){
                if(count0 > 0) hs0.add(count0);
                if(count1 > 0) hs1.add(count1);
            }
            
        }
        if( S.charAt(N-1) == '0' ) hs1.add(0);
        
        if( hs0.size() <= K ){
            System.out.println(N);
            return;
        }
        
        long maxSum = 0;
        for(int i = 0; i < K; i++){
            if(i == 0){
                maxSum += hs1.get(i) + hs0.get(i) + hs1.get(i+1);
            } else {
                maxSum += hs0.get(i) + hs1.get(i+1);
            }
        }
        
        long sum = maxSum;
        for(int i = 0; i < hs0.size()-K; i++){
            sum += hs0.get(K+i) + hs1.get(K+1+i);
            sum -= hs0.get(i) + hs1.get(i);
            maxSum = Math.max(maxSum, sum);
        }
        
        System.out.println(maxSum);
        
    }
}
