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: Accepted)

#include <bits/stdc++.h>

#define fr first
#define sc second
#define OK puts("OK");
#define pb push_back
#define mk make_pair

using namespace std;

typedef long long ll;

const int inf = (int)1e9 + 7;
const int N = (int)2e5 + 7;

int n,m;
int a[N],u[N],us[N],db[N],sn[N];
int ans;

 main () {
        cin >> n >> m;

        for (int i = 1; i <= n; i ++) {
                cin >> a[i];
                u[a[i] % m] ++;
                us[a[i]] ++;
        }
        for (int i = 1; i <= n; i ++) {
                if (us[a[i]]) {
                        db[a[i] % m] += (us[a[i]] - (us[a[i]] % 2));
                        sn[a[i] % m] += us[a[i]] % 2;
                        us[a[i]] = 0;
                }
        }
        for (int i = 1; i < (m + 1) / 2; i ++) {
                int f = u[i];
                int s = u[m - i];
                if (f < s) {
                        ans += f;
                        int p = min(0,sn[m - i] - f);
                        db[m - i] += p;
                        ans += (db[m - i] / 2);
                }
                else {
                        ans += s;
                        int p = min(0,sn[i] - s);
                        db[i] += p;
                        ans += (db[i] / 2);
                }
        }
        if (m % 2 == 0) {
                ans += (u[m / 2] / 2);
        }
        ans += (u[0] / 2);

        cout << ans << endl;
}