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.*;
import java.io.*;

public class Main {
	private static void solve(){
		//Implement solution here.
		int n = ni();
		int k = ni();
		char[] ca = next().toCharArray();
		int[] cnta = new int[n];

		int cnt = 1;
		int aidx = 0;
		for(int i = 1; i < ca.length; i++) {
			if(ca[i] != ca[i-1]) {				
				cnta[aidx] = cnt;
				aidx++;
				cnt = 1;
			} else {
				cnt++;
			}
		}
		cnta[aidx] = cnt;
//		Arrays.stream(cnta).forEach(v -> System.out.print(v +","));
//		System.out.println();	
		
		boolean isz =  ca[0] == '0';
		int tmp = 0;
		int result = 0;
		int scnt = 0;
		for(int i = 0; i < cnta.length; i++) {
			if(cnta[i] == 0) break;
			if(isz) {
				tmp += cnta[i];				
				if(scnt >= k) {
					tmp -= cnta[i - (2*k)];
					if(i - (2*k) > 0) tmp -= cnta[i - (2*k) - 1];
				}
				scnt++;
			}else {
				tmp += cnta[i];				
			}
			isz = !isz;
			result = Math.max(result, tmp);
		}
		System.out.println(result);
	}

	//Switch input source (stdin/file)
	private static BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args){
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		}
		solve();
	}

	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}