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
N, K = map(int, input().split())
S = list(input())
A = {}
if S[0] == "1":
    A[1] = 0
    c = 1
else:
    A[0] = 0
    c = 0
for i in range(N-1):
    if S[i] == "0" and S[i+1] == "1":
        c += 1
        A[c] = i+1
B = {}
c = 1
for i in range(N-1):
    if S[i] == "1" and S[i+1] == "0":
        B[c] = i
        c += 1

ans = 0
for p, v in A.items():
    t = B.get(p+K, N-1)
    ans = max(t-v+1, ans)
print(ans)