#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;

int main(){
  int n, m;
  cin >> n >> m;
  vector<int> num(m,0);
  vector<int> p(m,0);
  set<int> s;
  rep(i,n) {
    int x;
    cin >> x;
    num[x%m]++;
    auto itr = s.find(x);
    if(itr != s.end()) {
      p[x%m]++;
      s.erase(x);
    } else {
      s.insert(x);
    }
  }
  int ans = 0;
  ans += num[0]/2;
  if(m % 2 == 0) ans += num[m/2]/2;
  for(int i = 1; i < m-i; i++){
    int a = i, b = m - i;
    if(num[a] < num[b]) swap(a,b);
    ans += num[b];
    int c = num[a] - num[b];
    ans += min(c/2,p[a]);
  }
  cout << ans << endl;
  return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

7 5
3 1 4 1 5 9 2


Output

3


Input

15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99


Output

6

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;

int main(){
  int n, m;
  cin >> n >> m;
  vector<int> num(m,0);
  vector<int> p(m,0);
  set<int> s;
  rep(i,n) {
    int x;
    cin >> x;
    num[x%m]++;
    auto itr = s.find(x);
    if(itr != s.end()) {
      p[x%m]++;
      s.erase(x);
    } else {
      s.insert(x);
    }
  }
  int ans = 0;
  ans += num[0]/2;
  if(m % 2 == 0) ans += num[m/2]/2;
  for(int i = 1; i < m-i; i++){
    int a = i, b = m - i;
    if(num[a] < num[b]) swap(a,b);
    ans += num[b];
    int c = num[a] - num[b];
    ans += min(c/2,p[a]);
  }
  cout << ans << endl;
  return 0;
}
```