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()
c = [int(s[i]) for i in range(n)]
count = []
cost = 0
if c[0] == 0:
  cost += -1
else:
  cost += 1
for i in range(1,n):
  if c[i] == c[i-1]:
    cost += -1 if cost < 0 else 1
  else:
    count.append(cost)
    cost = -1 if cost > 0 else 1
count.append(cost)
i = 0
M = 0
l = 0
r = 0
for j in range(len(count)):
  if count[j] < 0:
    M += -count[j]
    i += 1
  else:
    M += count[j]
  if i == k:
    if j < len(count)-2:
      M += count[j+1]
      r = j + 1
    else:
      r = j
    break
tmp = M
if r < len(count) - 2:
  if count[0] < 0:
    tmp += count[0]
    l += 1
    r += 1
    tmp += -count[r]
    if r+1 < k:
      tmp += count[r + 1]
      r += 1
    M = max(M,tmp)
  while r + 1 < K -1:
    tmp -= count[l]
    tmp += count[l+1]
    l += 2
    r += 1
    tmp += -count[r]
    if r+1 < k:
      tmp += count[r + 1]
      r += 1
    M = max(M,tmp)
print(M)