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>
#include<iostream>
using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define UL unsigned long long


int main(void)
{
    IOS
    int n,k;
    cin >> n >> k;
    string s;
    cin >> s;

    vector<int> zero;
    vector<int> one;
    if (s[0] == '0') {
        one.emplace_back(0);
    }

    int cnt = 1;
    char c = s[0];
    for(auto i = 1; i < n; i++)
    {
        if (c == s[i]) {
            cnt++;
            //cerr << cnt;
        }
        if ((c != s[i]) || (i == n-1)) { 
            if (c == '0') {
                zero.emplace_back(cnt);
            } else {
                one.emplace_back(cnt);
            }
            c = s[i];
            cnt = 1;
        }
    }
    
    /*
    cerr << "zero: ";
    for(int i = 0; i < zero.size() ; i++) {
        cerr << zero[i] << " ";
    }
    cerr << "one: ";
    for(int i = 0; i < one.size() ; i++) {
        cerr << one[i] << " ";
    }
    */
    
    int m = zero.size();
    if (m < k) {
        cout << n << endl; 
        return 0;
    }

    int ans = 0;
    int count = 0;
    for(auto i = 0; i <= m-k+1; i++)
    {
        count += one[i] + zero[i];
        if (i >= k) {
            count -= one[i-k] + zero[i-k];
        }
        //cerr << "count: " << count << endl;
        if (i != m-k+1) {
            if (ans < count+one[i+k])  ans = count + one[i+k];
        } else {
            if (ans < count)  ans = count;
        }
    }
    

    cout << ans << endl;
    return 0;
}