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, raw_input().split())
S=raw_input()

c=S[0]
# T[0] := ("1" or "0", num)
T=[[c, 1]]
for s in S[1:]:
    if c==s:
        T[-1][1]+=1
    else:
        c=s
        T.append([c,1])

ans=-1
p=0
n=p
k=0
s=0
while n<len(T):
    if(T[n][0]=="0"):
        if k<K:
            k+=1
            s+=T[n][1]
        else:
            break
    else:
        s+=T[n][1]
    n+=1
ans=s
for pp in range(len(T)-1):
    p=pp+1
    if T[p][0]=="1":
        s-=T[p-1][1]
        if p+2*k-1<len(T):
            s+=T[p+2*k-1][1]
        if p+2*k<len(T):
            s+=T[p+2*k][1]
    else:
        s-=T[p-1][1]        
    ans=max(ans,s)  
print(ans)     

