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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
inline int_fast64_t read(){int_fast64_t x=0,f=1;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return x*f;}
void Main() {
    read();
    int K = read(), i = 1, ans = 0;
    vector<int> V;
    V.push_back(0);
    char c = getchar();
    while(c != '0' && c != '1'){
        c = getchar();
    }
    while(c == '0' || c == '1'){
        V.push_back(0);
        while (c == '1'){
            V[i]++;
            c = getchar();
        }
        i++;
        V.push_back(0);
        while (c == '0'){
            V[i]++;
            c = getchar();
        }
        i++;
    }
    int Vsize = V.size();
    for(int j = 1; j<Vsize; j++){
        V[j] += V[j-1];
    }
    for(int j = 1; j<Vsize; j += 2){
        ans = max(ans, V[min(j+2*K, Vsize-1)]-V[j-1]);
    }
    printf("%d\n", ans);
}
int main(){cin.tie(nullptr);ios::sync_with_stdio(false);Main();return 0;}