#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>

using namespace std;

#define mod 1000000007

int main()
{
    int n, m;
    cin >> n >> m;
    vector<vector<int> > amari(100001);
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        amari[x % m].push_back(x);
    }
    int ans = 0;
    for(int i = 0; i <= m / 2; i++){
        // cout << i << " " << ans << endl;
        if(i == 0){
            ans += amari[i].size() / 2;
        } else if(i == m / 2 && m % 2 == 0){
            ans += amari[i].size() / 2;
        } else {
            if(amari[i].size() > amari[m - i].size()) swap(amari[i], amari[m - i]);
            sort(amari[m - i].begin(), amari[m - i].end());
            while(amari[m - i].size() > amari[i].size()){
                bool canmakepair = false;
                for(int j = 0; j < amari[m - i].size() - 1; j++){
                    // cout << amari[m - i][j] << " ";
                    if(amari[m - i][j] == amari[m - i][j + 1]){
                        amari[m - i].erase(amari[m - i].begin() + j);
                        amari[m - i].erase(amari[m - i].begin() + j);
                        ans++;
                        canmakepair = true;
                        break;
                    }
                }
                // cout << endl;
                if(!canmakepair) break;
            }
            if(amari[m - i].size() < amari[i].size()) ans--;
            ans += amari[i].size();
        }
    }
    cout << ans << 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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>

using namespace std;

#define mod 1000000007

int main()
{
    int n, m;
    cin >> n >> m;
    vector<vector<int> > amari(100001);
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        amari[x % m].push_back(x);
    }
    int ans = 0;
    for(int i = 0; i <= m / 2; i++){
        // cout << i << " " << ans << endl;
        if(i == 0){
            ans += amari[i].size() / 2;
        } else if(i == m / 2 && m % 2 == 0){
            ans += amari[i].size() / 2;
        } else {
            if(amari[i].size() > amari[m - i].size()) swap(amari[i], amari[m - i]);
            sort(amari[m - i].begin(), amari[m - i].end());
            while(amari[m - i].size() > amari[i].size()){
                bool canmakepair = false;
                for(int j = 0; j < amari[m - i].size() - 1; j++){
                    // cout << amari[m - i][j] << " ";
                    if(amari[m - i][j] == amari[m - i][j + 1]){
                        amari[m - i].erase(amari[m - i].begin() + j);
                        amari[m - i].erase(amari[m - i].begin() + j);
                        ans++;
                        canmakepair = true;
                        break;
                    }
                }
                // cout << endl;
                if(!canmakepair) break;
            }
            if(amari[m - i].size() < amari[i].size()) ans--;
            ans += amari[i].size();
        }
    }
    cout << ans << endl;
    return 0;
}
```