#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <iostream>
#include <queue>
#include <list>
#include <stack>
#include <map>
#include <numeric>
#include <set>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define ALLOF(c) (c).begin(), (c).end()
typedef long long ll;

int main(){
  ios::sync_with_stdio(false);
  ll N, M;
  cin >> N >> M;
  map<ll, vector<ll> > memo;
  rep(i,N){
    ll a;
    cin >> a;
    memo[a%M].push_back(a);
  }


  ll ret = 0;
  rep(i,M){
    if(i > M-i) break;
    if(i==0 || (M%2==0 && i==M/2)){
      vector<ll> a = memo[i];
      ret += a.size()/2;
    }else{
      vector<ll> a = memo[i];
      vector<ll> b = memo[M-i];
      if(a.size() > b.size()) swap(a,b);
      map<ll,ll> cnt;
      rep(j,b.size()){
        cnt[b[j]]++;
      }
      vector<ll> tmp;
      FOR(it,cnt){
        tmp.push_back(it->second);
      }

      ll m = b.size();
      rep(j,tmp.size()){
        while(tmp[j] >= 2 && m-2 >= a.size()){
          m -= 2;
          tmp[j] -= 2;
          ret++;
        }
      }
      ret += a.size();
    }
  }

  cout << ret << endl;
  
  return 0;
}


### Prompt

Please provide a CPP coded solution to the problem described below:

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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <iostream>
#include <queue>
#include <list>
#include <stack>
#include <map>
#include <numeric>
#include <set>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
#define REP(i,a,n) for(int i=(a); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin(); it!=(c).end(); ++it)
#define ALLOF(c) (c).begin(), (c).end()
typedef long long ll;

int main(){
  ios::sync_with_stdio(false);
  ll N, M;
  cin >> N >> M;
  map<ll, vector<ll> > memo;
  rep(i,N){
    ll a;
    cin >> a;
    memo[a%M].push_back(a);
  }


  ll ret = 0;
  rep(i,M){
    if(i > M-i) break;
    if(i==0 || (M%2==0 && i==M/2)){
      vector<ll> a = memo[i];
      ret += a.size()/2;
    }else{
      vector<ll> a = memo[i];
      vector<ll> b = memo[M-i];
      if(a.size() > b.size()) swap(a,b);
      map<ll,ll> cnt;
      rep(j,b.size()){
        cnt[b[j]]++;
      }
      vector<ll> tmp;
      FOR(it,cnt){
        tmp.push_back(it->second);
      }

      ll m = b.size();
      rep(j,tmp.size()){
        while(tmp[j] >= 2 && m-2 >= a.size()){
          m -= 2;
          tmp[j] -= 2;
          ret++;
        }
      }
      ret += a.size();
    }
  }

  cout << ret << endl;
  
  return 0;
}


```