#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
	int n, m;
	cin >> n >> m;
	vector<vector<int>> vv(m);
	for(int i = 0; i < n; i++){
		int x;
		cin >> x;
		vv[x % m].emplace_back(x);
	}
	for(int i = 0; i < m; i++) sort(vv[i].begin(), vv[i].end());
	int ans = 0;
	for(int i = 0; i < m; i++){
		int j = (m - i) % m;
		if(i > j) continue;
		if(i == j){
			ans +=vv[i].size() / 2;
			continue;
		}
		int p = 0;
		for(int k = 0; k + 1 < vv[i].size(); k++) if(vv[i][k] == vv[i][k + 1]){
			p++;
			k++;
		}
		int q = 0;
		for(int k = 0; k + 1 < vv[j].size(); k++) if(vv[j][k] == vv[j][k + 1]){
			q++;
			k++;
		}
		int is = vv[i].size(), js = vv[j].size();
		if(is > js){
			swap(is, js);
			swap(p, q);
		}
		int sel = min((js - is) / 2, q);
		ans += sel;
		js -= sel * 2;
		ans += min(is, js);
	}
	cout << ans << endl;
}
### 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
	int n, m;
	cin >> n >> m;
	vector<vector<int>> vv(m);
	for(int i = 0; i < n; i++){
		int x;
		cin >> x;
		vv[x % m].emplace_back(x);
	}
	for(int i = 0; i < m; i++) sort(vv[i].begin(), vv[i].end());
	int ans = 0;
	for(int i = 0; i < m; i++){
		int j = (m - i) % m;
		if(i > j) continue;
		if(i == j){
			ans +=vv[i].size() / 2;
			continue;
		}
		int p = 0;
		for(int k = 0; k + 1 < vv[i].size(); k++) if(vv[i][k] == vv[i][k + 1]){
			p++;
			k++;
		}
		int q = 0;
		for(int k = 0; k + 1 < vv[j].size(); k++) if(vv[j][k] == vv[j][k + 1]){
			q++;
			k++;
		}
		int is = vv[i].size(), js = vv[j].size();
		if(is > js){
			swap(is, js);
			swap(p, q);
		}
		int sel = min((js - is) / 2, q);
		ans += sel;
		js -= sel * 2;
		ans += min(is, js);
	}
	cout << ans << endl;
}
```