Problem:
Score : 700 points

Problem Statement

Takahashi is playing with N cards.

The i-th card has an integer X_i on it.

Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:

The integers on the two cards are the same.

The sum of the integers on the two cards is a multiple of M.

Find the maximum number of pairs that can be created.

Note that a card cannot be used in more than one pair.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦X_i≦10^5

Input

The input is given from Standard Input in the following format:

N M
X_1 X_2 ... X_N

Output

Print the maximum number of pairs that can be created.

Sample Input 1

7 5
3 1 4 1 5 9 2

Sample Output 1

3

Three pairs (3,2), (1,4) and (1,9) can be created.

It is possible to create pairs (3,2) and (1,1), but the number of pairs is not maximized with this.

Sample Input 2

15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99

Sample Output 2

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

#define pb push_back
#define mp make_pair

int n, m;
int mas[100000] = {};
int cnt[100000] = {};
int p[100000] = {};

int main()
{
        std::cin.sync_with_stdio(0);
        cin >> n >> m;
        for(int i = 0; i < n; i++){
                int t;
                cin >> t;
                cnt[t]++;
                if(cnt[t] % 2 == 0) p[t % m]++;
                t %= m;
                mas[t]++;
        }

        int ans = 0;
        for(int i = 0; i <= m / 2; i++){
                int cur = i;
                int cur2;
                if(i == 0) cur2 = 0;
                else cur2 = m - i;

                if(cur == cur2){
                        int mx = mas[cur] / 2;
                        ans += mx;
                }
                else{
                        int mx = min(mas[cur], mas[cur2]);
                        mas[cur] -= mx;
                        mas[cur2] -= mx;

                        int probPairs = mas[cur] / 2;
                        p[cur] = min(p[cur], probPairs);

                        probPairs = mas[cur2] / 2;
                        p[cur2] = min(p[cur2], probPairs);

                        ans += mx + p[cur] + p[cur2];
                }
        }
        cout << ans;

        return 0;
}