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=map(int, input().split())
s=input()
t=[]
now=s[0]
count=0
for i in s:
    if now==i:
        count+=1
    else:
        t.append(count)
        now=i
        count=1
t.append(count)
if k>len(t)//2:print(sum(t))
else:
    ans=0
    if s[0]=='0':
        ans=max(ans,t[2*k])
        t=t[1:]
    if s[-1]=='0':
        ans=max(ans,t[-2*k:-1])
        t=t[:-2]
    ans2=sum(t[:2*k+1])
    ans=max(ans,ans2)
    for i in range(len(t)//2-k):
        ans2=ans2-sum(t[2*i:2*i+2])+sum(t[2*i+2*k+1: 2*i+2*k+1+2])
        ans=max(ans,ans2)
print(ans)