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: Accepted)

data = input().split()
N = int(data[0])
K = int(data[1])
 
TL = list(input())
TLlist = []
TLLlist = []
now = TL[0]
nowlist = now
anslist = []
 
for lop in range(1,len(TL)):
    if not (now == TL[lop]):
        TLlist.append(nowlist)
        TLLlist.append(len(nowlist))
        nowlist = TL[lop]
        now = TL[lop]
    else:
        nowlist += now
TLlist.append(nowlist)
TLLlist.append(len(nowlist))
 
 
if TLlist[0][:1] == "0" :
    anslist.append( sum(TLLlist[:2*K]) )
    anslist.append( sum(TLLlist[1:2*K + 2]) )
    for lop in range(3,len(TLlist),2):
        if 2*K +  lop > len(TLlist):
            break
        anslist.append( anslist[len(anslist)-1] - sum(TLLlist[lop-2:lop]) + sum(TLLlist[lop + 2*K - 1 : lop + 2*K + 1])  )
elif TLlist[0][:1] == "1" :
    anslist.append( sum(TLLlist[:2*K + 1 ]) )
    for lop in range(2,len(TLlist),2):
        if 2*K +  lop > len(TLlist):
            break
        anslist.append( anslist[len(anslist)-1] - sum(TLLlist[lop-2:lop]) + sum(TLLlist[lop + 2*K - 1 : lop + 2*K + 1])  )
 
print(max(anslist))