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 = input().split()
N = int(N)
K = int(K)
S = input()
A = []
for i in range(len(S)-1):
    if (S[i]=="0" and S[i+1]=="1") or (S[i]=="1" and S[i+1]=="0"):
        A.append(i+1)
A.insert(0, 0)
A.append(N)
if K > len(A)//2:
    K = len(A)//2
M = []
while K > 0:
    if S[0] == "0":
        M.append(A[K*2]-A[0])
        a = 1
        b = a + K*2 + 1
    elif S[0] == "1":
        a = 0
        b = a + K*2 + 1
    if S[-1] == "1":
        while b <= len(A):
            M.append(A[b]-A[a])
            a += 2
            b = a + K*2 + 1
    elif S[-1] == "0":
        while b <= len(A)-1:
            M.append(A[b]-A[a])
            a += 2
            b = a + K*2 + 1
        M.append(A[-1]-A[-1-K*2])
    K -= 1
print(max(M))