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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
using namespace std;
using Graph = vector<vector<int>>;
typedef long long ll;
typedef pair<int, int> P;
const int inf = 1 << 21;
const ll INF = 1LL << 60;
const ll mod = 1e9+7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// abc124d
int main(){
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    vector<int> ans;
    int pre = 1;
    if (s[0]=='0'){
        ans.push_back(0);
        ans.push_back(1);
        pre = 0;
    }else{
        ans.push_back(1);
    }
    rep(i, n){
        if (i==0) continue;
        if (pre == s[i] - '0'){
            ans[ans.size()-1] = ans[ans.size()-1]+1;
        }else{
            ans.push_back(1);
            pre = s[i]-'0';
        }
    }
    if (pre==0) ans.push_back(0);
    rep(i, ans.size()-1) ans[i+1] = ans[i]+ans[i+1];
    int ansm = 0;
    if (2*k+1>=n){
        cout << n << endl;
        return 0;
    }else{
        rep(i, n){
            int l = i, r = 2*k+i;
            if (i%2==1) continue;
          	if (r >= ans.size()) break;
            if (l==0){
                chmax(ansm, ans[r]);
            }else{
                chmax(ansm, ans[r]-ans[l]);
            }
        }
    }
    cout << ansm << endl;
    return 0;
}