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: Wrong Answer)



N, K = map(int, input().split())
S = input()

ones = []
t = 0
for i in range(N):
    if i == 0:
        t += 1
        pass
    elif (S[i - 1] != S[i]):
        ones.append(t)
        t = 1
    else:
        t += 1
ones.append(t)

if S[0] == '0':
    t = ones
    ones = [0]
    ones.extend(t)

ones1 = []
for k in range(0, len(ones), 2):
    temp_k = 0
    if k < len(ones):
        temp_k += ones[k]
    if (k + 1) < len(ones):
        temp_k += ones[k + 1]
    if (k + 2) < len(ones):
        temp_k += ones[k + 2]
    ones1.append(temp_k)


f = True
res = 0
for k in range(K):
    if k >= len(ones1):
        print(res)
        f = False
        break
    res += ones1[k]

temp = 0
if f:
    for k in range(k, len(ones1), 2):

        temp_k = 0
        if k < len(ones):
            temp_k += ones[k] - 1
        if (k + 1) < len(ones):
            temp_k += ones[k + 1] - 1

        temp_res = res - temp + temp_k
        temp = temp_k
        if res < temp_res:
            res = temp_res

    print(res)