Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in Python (3.4.3) (judge verdict: Runtime Error)

N,K = map(int,input().split())
S = input()
S_0 = list(S.split("1"))
S_1 = list(S.split("0"))

#print(S_0,S_1)

S_1_len = [len(i) for i in S_1]
S_1_len = [j for j in S_1_len if j != 0]

S_0_len = [len(i) for i in S_0]
S_0_len = [j for j in S_0_len if j != 0]

#print(S_1_len)
#print(S_0_len)
if S_0_len:
    max0_index = S_0_len.index(max(S_0_len))
else:
    max0_index = 0

#print(max0_index)
count1 = 0

if S[0] == "1":
        #S_1_len[max0_index] = S_0_len[max0_index] + S_1_len[max0_index] + S_1_len[max0_index+1]
    count += (S_0_len[max0_index] + S_1_len[max0_index] + S_1_len[max0_index+1])
    for i in range(K-1):
        count += (S_0_len[max0_index-i] + S_1_len[max0_index-i])
else:
    count += (S_0_len[max0_index] + S_1_len[max0_index] + S_1_len[max0_index-1])
    for i in range(K-1):
        count += (S_0_len[mex0_index-i] + S_1_len[max0_index-i])

print(count)