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 <cmath>
#include <vector>
#include <map>
#include <functional>
#include <queue>
#include <iostream>
#include <string.h>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <cstdint>
#include <climits>
#include <unordered_set>
#include <sstream>

using namespace std;

#define ll long long int

struct UnionFind {
    vector<int> data;
    UnionFind(int size) : data(size, -1)
    {
    }
    bool unionSet(int x, int y) {
        x = root(x); y = root(y);
        if (x != y) {
            if (data[y] < data[x]) swap(x, y);
            data[x] += data[y]; data[y] = x;
        }
        return x != y;
    }
    bool findSet(int x, int y) {
        return root(x) == root(y);
    }
    int root(int x) {
        return data[x] < 0 ? x : data[x] = root(data[x]);
    }
    int size(int x) {
        return -data[root(x)];
    }
};

ll calc(int a, int b, int a2, int b2)
{
    ll sum = 0;
    sum += min(a,b);
    a = a - sum;
    b = b - sum;
    sum += min(a/2,a2);
    sum += min(b/2,b2);
    return sum;
}

int main()
{
    int n, m;
    cin >> n >> m;
    map<int,int> v1;
    map<int,int> v;
    for(int i = 0;i < n;i++)
    {
        int a;
        cin >> a;
        v1[a]++;
        v[a % m]++;
    }
    map<int,int> t;
    for(auto r : v1)
    {
        t[r.first % m] += v1[r.second / 2];
    }
    
    for(int i = 0;i < m;i++)
    {
        //cout << v[i] << "," << t[i] << endl;
    }
    
    ll sum = 0;
    sum += v[0]/2;
    //cout << endl << sum << endl;
    if(m % 2 == 0)
    {
        sum += v[m/2]/2;
        for(int i = 1; i < m/2;i++)
        {
            int f = v[i];
            int o = v[m-i];
            ll c = calc(f,o, t[i], t[m-i]);
            //cout << i << "," << c << endl;
            sum += c;
        }
    }
    else
    {
        for(int i = 1; i <= m/2;i++)
        {
            int f = v[i];
            int o = v[m-i];
            ll c = calc(f,o, t[i], t[m-i]);
            //cout << i << "," << c << endl;
            sum += c;
        }
    }
    cout << sum << endl;
    return 0;
}