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)

import numpy as np

N,K = map(int, input().split())
s = list(input())
S = np.zeros(len(s)+1)
sl = []
for i in range(len(s)):
    S[i] = int(s[i])

S[len(s)] = 1 - S[len(s)-1]

#print(S)
a = 1
for i in range(1,len(s)+1):
    
    if S[i] != S[i-1]:
        
        sl.append(a)
        #print(i,a)
        a = 1
    else :
        a += 1

slm = [] 
for i in range (len(sl) - (2*K)):
    #print(i,K)
    #print(sl[0:2])
    if i == 0:
        slm.append(sum(sl[i:i+2*K]))
    else:
        slm.append(sum(sl[i:i+2*K+1]))
if len(sl) - (2*K) > 0:
    print(max(slm))
else:
    print(max(sl))