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.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
n,k=map(int,input().split())
s=input()
b=[1]
if s[0]=="0":b=[0,1]
for i in range(1,n):
  if s[i-1]!=s[i]:b+=[1]
  else:b[-1]+=1
if s[-1]=="0":b+=[0]
if len(b)//2<=k:print(n);exit()
#print(b)

x=a=sum(b[:k*2+1])
#print(b[:k*2+1])
for i in range(k,len(b)//2):
  a+=b[i*2+1]+b[(i+1)*2]
  a-=b[(i-k)*2]+b[(i-k)*2+1]
  x=max(a,x)
  #print(b[(i-k)*2+2:(i+1)*2+1])
print(x)