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.io.*;
import java.util.ArrayList;

class Main{
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String str = br.readLine();
        String[] sl = str.split(" ");
        int N = Integer.parseInt(sl[0]);
        int K = Integer.parseInt(sl[1]);
        str = br.readLine();
        ArrayList<Integer> nums = new ArrayList<>();
        int tmp = 1;
        int cont = 0;
        for (String s:str.split("")){
            if (Integer.parseInt(s)== tmp){
                cont++;
            } else {
                nums.add(cont);
                cont = 1;
                tmp = 1 - tmp;
            }
        }
        if (cont != 0){
            nums.add(cont);
        }
        if (nums.size() % 2 == 0){
            nums.add(0);
        }
        Integer[] nums_ = nums.toArray(new Integer[nums.size()]);
        ArrayList<Integer> ans = new ArrayList<>();
        ArrayList<Integer> cml = new ArrayList<>();
        cml.add(0);
        for (int i=0;i<nums_.length;i++){
            cml.add(cml.get(i)+nums_[i]);
        }
        int k = K * 2 + 1;
        for (int i=0;i<cml.size();i+=2){
            ans.add(cml.get(Math.min(i+k,cml.size()-1))-cml.get(i));
        }
        int max = 0;
        for (int i:ans){
            if (i > max){
                max = i;
            }
        }
        System.out.println(max);
    }
}