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 = [int(i) for i in input().split()]
S = input()

si = [1]
si.extend(2+i for i, (s1, s2) in enumerate(zip(S, S[1:])) if s1 != s2)
si.append(N + 1)


def f(i, si):
    return si[min(i + 2*K + 1, len(si)) - 1] - si[i-1]

j0 = min(1 + 2*K + (1 if S[0] == "1" else 0), len(si)) - 1
x0 = si[j0] - 1

x = [x0]
x.extend(f(i+1, si) for i in range(2 if S[0] == "1" else 1, len(si), 2))
print(max(x)) 