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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int main(){
  int N, M;
  cin >> N >> M;
  vector<int> X(N);
  for (int i = 0; i < N; i++){
    cin >> X[i];
  }
  vector<map<int, int>> mp(M);
  for (int i = 0; i < N; i++){
    mp[X[i] % M][X[i]]++;
  }
  int ans = 0;
  for (int i = 0; i <= M / 2; i++){
    if (i == 0 || i * 2 == M){
      int tmp = 0;
      for (auto P : mp[i]){
        tmp += P.second;
      }
      ans += tmp / 2;
    } else {
      int sum1 = 0;
      for (auto P : mp[i]){
        sum1 += P.second;
      }
      int sum2 = 0;
      for (auto P : mp[M - i]){
        sum2 += P.second;
      }
      ans += min(sum1, sum2);
      int d = abs(sum1 - sum2);
      vector<int> val;
      if (sum1 > sum2){
        for (auto P : mp[i]){
          val.push_back(P.second);
        }
      } else {
        for (auto P : mp[M - i]){
          val.push_back(P.second);
        }
      }
      int sum = max(sum1, sum2);
      int cnt = val.size();
      int odd_cnt = 0;
      for (int i = 0; i < cnt; i++){
        if (val[i] % 2 == 1){
          odd_cnt++;
        }
      }
      if (odd_cnt < min(sum1, sum2)){
        ans += d / 2;
      } else {
        ans += (d - (odd_cnt - d) / 2) / 2;
      }
    }
  }
  cout << ans << endl;
}