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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define fbo find_by_order
#define ook order_of_key

typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef long double ld;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;

long long INF=numeric_limits<long long>::max();

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
    ll n,m,ans(0);
    cin >> n >> m;
    ll mod[m],same[100001],smod[m];
    memset(mod,0,sizeof(mod));
    memset(smod,0,sizeof(smod));
    memset(same,0,sizeof(same));
    vector< ll > x(n);
    for(ll i = 0; i < n; i++)
    {
        cin >> x[i];
        same[x[i]]++;
    }
    for(ll i = 1; i <= 100000; i++)
    {
        ll a = i%m;
        mod[a]+=same[i];
        smod[a]+=same[i]/2;
    }
    ans += mod[0]/2;
    if(m%2==0)ans += mod[m/2]/2;
    for(ll i = 1; i < (m+1)/2; i++)
    {
        ll j = m-i;
        //cout << mod[i] << " " << mod[j] << endl;
        if(mod[i]==mod[j])
        {
            ans += mod[i];
        }
        else if(mod[i]>mod[j])
        {
            ans += mod[j];
            ll left = mod[i] - mod[j];
            //cout << left/2 << " " << smod[i] << endl;
            left = min(left/2,smod[i]);
            ans += left;
        }
        else
        {
            ans += mod[i];
            ll left = mod[j] - mod[i];
            left = min(left/2,smod[j]);
            ans += left;
        }
    }
    cout << ans << endl;
}