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;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for(int i = 0; i < n; i++)

int main(void){
    ll n, k;
    cin >> n >> k;

    string s;
    cin >> s;

    vector<ll> count;
    ll tmp = 0;

    if(s[0] == '0'){
        count.push_back(0);
    }
    rep(i, n){
        if(i == 0){
            tmp = 1;
        }else if(s[i] == s[i-1]){
            tmp += 1;
        }else{
            count.push_back(tmp);
            tmp = 1;
        }
    }
    count.push_back(tmp);
    if(s[s.size()-1] == '0'){
        count.push_back(0);
    }

    rep(i, count.size()){
        if(i != 0){
            count[i] += count[i-1];
        }
    }

    ll ma = 0;
    for(int i = k * 2; i < count.size(); i += 2){
        ll res = count[i];
        if(i - k * 2 - 1 >= 0){
            res = count[i] - count[i - k * 2 - 1];
        }
        ma = max(ma, res);
    }

    if(k * 2 >= count.size()){
        cout << s.size() << endl;
    }else{
        cout << ma << endl;
    }
    return 0;


}