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: Time Limit Exceeded)

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <utility>
#include <map>
#include <bitset>
#include <numeric>
#define ll long long
using namespace std;
using p = pair<ll, ll>;
ll dx[8]={ 0, 1, 0,-1, 1, 1,-1,-1}; // x軸方向への変位
ll dy[8]={ 1, 0,-1, 0, 1,-1, 1,-1}; // y軸方向への変位

int main(void) {
    ll n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    s += 'W';
    vector<ll> a;
    a.push_back(0);
    for(ll i = 0; i < s.length(); i++){
        if(s[i]=='0' && s[i+1]=='1') a.push_back(i+1);
    }
    ll ans = 0;
    ll count = 0;
    bool ok = false;
    for(ll i = 0; i < a.size(); i++){
        count = 0;
        if(ok==true) break;
        for(ll j = a[i]; j < n; j++){
            if(s[j]=='0' && s[j+1]=='1') count++;
            if(count==k && s[j+1]=='0' ||  j==n-1) {
                ans = max(ans, j-a[i]+1);
                if(j==n-1) ok = true;
                break;
            }
        }
    }
    cout << ans << endl;
    return 0;
}