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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define llong long long
#define INF (__INT32_MAX__ / 2)
#define EPS 10e-8
#define MAX_NMX 100000

using namespace std;
typedef pair<llong, llong> ipair;
llong modulo(llong left, llong right) { return (left % right + right) % right; }

int N, M, X[MAX_NMX];

int numbers[MAX_NMX + 1];
int not_paired[MAX_NMX];
int pairs[MAX_NMX];


int solve() {
    for(int i = 0; i < N; ++i)
        numbers[X[i]]++;

    for(int i = 0; i <= MAX_NMX; ++i) {
        pairs[i % M] += numbers[i] / 2;
        if(numbers[i] % 2 == 1)
            not_paired[i % M]++;
    }

    int res = 0;
    for(int i = 0; i * 2 <= M; ++i)
        if(i == 0 || i * 2 == M) {
            res += not_paired[i] / 2;
        } else {
            
            if(not_paired[i] > not_paired[M - i]) { 
                int carry = min(not_paired[i] - not_paired[M - i], 2 * pairs[M - i]);
                res += not_paired[M - i] + carry;
                pairs[M - i] = (2 * pairs[M - i] - carry) / 2;

            } else if(not_paired[i] < not_paired[M - i]) { 
                int carry = min(not_paired[M - i] - not_paired[i], 2 * pairs[i]);
                res += not_paired[i] + carry;
                pairs[i] = (2 * pairs[i] - carry) / 2;

            } else {
                res += not_paired[i];
            }
        }

    for(int i = 0; i < M; ++i)
        res += pairs[i];

    return res;
}

int main(int argc, char **argv) {
    cin >> N >> M;
    for(int i = 0; i < N; ++i)
        cin >> X[i];

    cout << solve() << endl;   
}