#include <bits/stdc++.h>
using namespace std;

int main() {
    //ifstream cin("testD.in");
    int n, m; cin >> n >> m;
    
    vector<vector<int>> r(m);

    for(int i = 0; i < n; ++i) {
        int x; cin >> x;
        r[x % m].push_back(x);
    }
    
    int ans = 0;
    
    vector<int> taken(m, 0);
    
    for(int i = 1; i <= m / 2 - (1 - m % 2); ++i) {
        int temp = min(r[i].size(), r[m - i].size());
        taken[i] = taken[m - i] = temp;
        ans += temp;
    }
    
    for(int i = 0; i < m; ++i) {
        if(i == 0 || (m % 2 == 0 && i == m / 2)) {
            ans += r[i].size() / 2;
            continue;
        }

        map<int, int> f;
        for(auto temp : r[i])
            f[temp]++;

        int all = 0, singles = 0;

        for(auto temp : f)
            if(temp.second >= 2) {
                all += temp.second - temp.second % 2;
                singles += temp.second % 2;
            } else {
                singles++;
            }

        int rem = max(0, (taken[i] - singles));

        
        ans += (all - rem) / 2;
    }

    cout << ans << "\n";
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;

int main() {
    //ifstream cin("testD.in");
    int n, m; cin >> n >> m;
    
    vector<vector<int>> r(m);

    for(int i = 0; i < n; ++i) {
        int x; cin >> x;
        r[x % m].push_back(x);
    }
    
    int ans = 0;
    
    vector<int> taken(m, 0);
    
    for(int i = 1; i <= m / 2 - (1 - m % 2); ++i) {
        int temp = min(r[i].size(), r[m - i].size());
        taken[i] = taken[m - i] = temp;
        ans += temp;
    }
    
    for(int i = 0; i < m; ++i) {
        if(i == 0 || (m % 2 == 0 && i == m / 2)) {
            ans += r[i].size() / 2;
            continue;
        }

        map<int, int> f;
        for(auto temp : r[i])
            f[temp]++;

        int all = 0, singles = 0;

        for(auto temp : f)
            if(temp.second >= 2) {
                all += temp.second - temp.second % 2;
                singles += temp.second % 2;
            } else {
                singles++;
            }

        int rem = max(0, (taken[i] - singles));

        
        ans += (all - rem) / 2;
    }

    cout << ans << "\n";
}
```