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 <iostream>
#include <iomanip>
#include <cstdio>
#include <algorithm>
#include <numeric>
#include <random>
#include <vector>
#include <array>
#include <bitset>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>

using namespace std;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n==0) ? 1 : 10*TEN(n-1); }
int bsr(int x) { return 31 - __builtin_clz(x); }

const int MN = 100100;
int main() {
    cout << setprecision(20);
    int n, m;
    cin >> n >> m;
    int co[MN] = {};
    int ok[MN] = {};
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        co[x % m]++;
        mp[x]++;
    }
    for (auto p: mp) {
        ok[p.second % m] += p.first/2;
    }
    int sm = 0;
    sm += co[0]/2;
    if (m % 2 == 0) sm += co[m/2]/2;
    for (int l = 1; l < m; l++) {
        int r = m-l;
        if (l >= r) break;
        sm += min(co[l], co[r]);
//        cout << co[l] << " " << co[r] << " d " << l << endl;
        if (co[l] < co[r]) {
            int y = (co[r]-co[l])/2;
            sm += min(y, ok[r]);
        } else {
            int y = (co[l]-co[r])/2;
            sm += min(y, ok[l]);
        }
    }
    cout << sm << endl;
    return 0;
}