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) {		
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		String S = sc.next();
		sc.close();
		
		int ones[] = new int[N];
		int zeros[] = new int[N];
		Arrays.fill(ones,-1); Arrays.fill(zeros,-1);
		
		char before = '1';
		boolean zeroStart = false;
		if(S.charAt(0)=='0') {
			before = '0';
			zeroStart = true;
		}
		
		char Slast = S.charAt(N-1);
		
		int oneP = 0 , zeroP = 0;
		
		int cnt = 0;
		for(int i=0;i<N;i++) {
			char s = S.charAt(i);
			if(s == before) {				
				cnt ++;
			}else {
				if(s=='0') {
					ones[oneP] = cnt;
					oneP++;
					cnt = 1;
					before = '0';
				}else {
					zeros[zeroP]=cnt;
					zeroP++;
					cnt = 1;
					before = '1';
				}
			}
		}
		if(S.charAt(N-1)=='0') {
			zeros[zeroP] = cnt;
		}else {
			ones[oneP] = cnt;
		}
	
		
		if(ones[0]==-1) {
			System.out.println(zeros[0]);
			return;
		}
		
		int ans = 0;
		boolean fin = false;
		if(!zeroStart) {
			for(int i=0;i<N;i++) {
				int tmp = ones[i];
				int last = Math.min(K,N-1);
				for(int j = 0;j<last;j++) {
						if(ones[i+j+1]<0||zeros[i+j]<0) {
							if(Slast=='0') {
								tmp += zeros[i+j];
							}
							fin = true;
							break;
						}
						tmp += ones[i+j+1]+zeros[i+j];
				}
				ans = Math.max(ans,tmp);
				if(fin)break;
			}
		}
		else {
			for(int i=0;i<N;i++) {
				int tmp = ones[i];
				int last = Math.min(K,N-1);
				for(int j = 0;j<last;j++) {
					if(ones[i+j+1]<0||zeros[i+j+1]<0) {
						if(Slast=='0') {
							tmp += zeros[i+j+1];
						}
						fin = true;
						break;
					}
					tmp += ones[i+j+1]+zeros[i+j+1]; 
				}
				ans = Math.max(ans,tmp);
				if(fin)break;
			}
			int t = 0;
			for(int i=0;i<K;i++) {
				if(ones[i]<0||zeros[i]<0) break;
				t += zeros[i]+ones[i];
			}
			ans = Math.max(ans,t);
		}
		
		System.out.println(ans);
	}

	
}