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()
def parse(s):
    arr = []
    prev = -1
    ans = 0
    for i in range(n):
        if s[i] == '0':
            if  prev == 0:
                ans += 1
            else:
                arr.append([prev,ans])
                ans = 1
                prev = 0
        elif s[i] == '1':
            if prev == 1:
                ans += 1
            else:
                arr.append([prev,ans])
                ans = 1
                prev = 1
    arr.append([prev,ans ])
    return arr
get = parse(s)[1:]


ll = len(get)

# print(get)

wL = wR = 0
bestL = bestWindow = 0
zeroCount = 0
ans = 0
while wR < ll:
    if zeroCount <= k :
        if get[wR][0] == 0 :
            zeroCount += 1
            ans += get[wR][1]
        else:
            ans += get[wR][1]
        wR += 1
    if zeroCount > k :
        if get[wL][0] == 0 :
            zeroCount -= 1
            ans -= get[wL][1]
        else:
            ans -= get[wL][1]

        wL += 1
    if (ans > bestWindow) and (zeroCount<=k) :
        bestWindow = ans
        bestL = wL

print(bestWindow)