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>
using namespace std;

using P = pair<int,int>;
using P3 = pair<P, int>;
using ll = long long int;
constexpr int INF = 1<<30;

int main() {
    int n, k;
    string s;
    cin >> n >> k >> s;
    s.push_back('2');
    vector<int> a;
    int cnt = 1;
    for(int i=0;i<n;i++){
        if(s[i] != s[i+1]){
            a.push_back(cnt*(s[i]=='1'?1:-1));
            cnt = 1;
        }
        else{
            cnt++;
        }
    }
    int ans = 0, len = 0, m = a.size(), x = 0;
    for(int r=0,l=0; r<m;){
        while(l<r && x > k){
            len -= abs(a[l]);
            if(a[l] < 0) x--;
            l++;
        }
        if(x <= k) ans = max(ans, len);
        while(r<m && x <= k){
            len += abs(a[r]);
            if(a[r] < 0) x++;
            if(x <= k) ans = max(ans, len);
            r++;
        }
    }
    cout << ans << endl;
    return 0;
}