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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
long long mo = 1e9 + 7;
typedef long long ll;
typedef pair<int,int> P;

int main(){
    ll N,K;
    cin >> N >> K;
    string S;
    cin >> S;
    vector<ll> v;
    ll now = 1;
    ll cnt = 0;
    rep(i,S.size()){
        if(S[i] == char('0'+now)){
            cnt++;
        }else{
            v.push_back(cnt);
            now = 1-now;
            cnt = 1;
        }
    }
    //if(cnt != 0)
    v.push_back(cnt);
    if(v.size()%2 == 0){
        v.push_back(0);
    }
    
    if(v[0] == N){
        cout << N << endl;
        return 0;
    }
    vector<ll> memo;
    rep(i,v.size()-(2*K+1) + 1){
        ll sum = 0;
        rep(j,2*K+1){
            if(i+j >= v.size())break;
            sum += v[i+j];
            //sum += v[0];
        }
        memo.push_back(sum);// i = (v.size() -2*K -1)     j=(2*K)
        i++;
    }
    ll M = 0;
    rep(i,memo.size()){
        M = max(M,memo[i]);
    }
    //cout << (1^0) << endl;
    /*
    rep(i,v.size()){
        cout << v[i] << " ";
    }
    cout << endl;*/
    cout << M << endl;
}