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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.next());
		int K = Integer.parseInt(sc.next());
		String S = sc.next();
		sc.close();

		int[] cntZero = new int[N+2];
		int[] cntOne = new int[N];
		boolean flg = false;
		int numZ = 0;
		int numO = 0;
		if(S.charAt(0)=='0') {
			flg=true;
			cntOne[numO++]=0;
		}
		for(int i =0;i < N;i++) {
			if(!flg && S.charAt(i)=='1') {
				cntOne[numO++] = i;
				flg = true;
			}else if(flg && S.charAt(i)=='0'){
				cntZero[numZ++] = i;
				flg = false;
			}
		}
		cntZero[numZ++] = N;
		int max = 0;
		if(numZ -1 < K) {
			max = N;
		}else{
			for(int i = K;i < numZ;i++) {
				if(cntZero[i] - cntOne[i-K] > max) {
					max = cntZero[i] - cntOne[i-K];
				}
			}
		}
		System.out.println(max);
	}
}