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
size,K = map(int, raw_input().split())
S = raw_input()
flag = 0
a = []
temp1 = []
temp2 = []
n = 0
for c in S:
	if(flag == 0):
		if(c == '1'):
			temp1.append(n)
			flag = 1
	else:
		if(c == '0'):
			flag = 0
			temp2.append(n)
	n += 1
n = 0
if(K <= len(temp2)):
	a.append(temp2[K-1])
	for i in temp1:
		if(n + K < len(temp2)):
			a.append(temp2[n + K] - i)
		else:
			a.append(size - i)
		n += 1
else:
	a.append(size)
print(max(a))