#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100;
vector<int> e[maxn];
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < n; i++){
		int x; scanf("%d", &x);
		e[x % m].push_back(x);
	}
	int ans = e[0].size() / 2;
	for(int i = 1; i < m; i++){
		int j = m - i;
		if(i == j){
			ans += e[i].size() / 2;
		} else if(i < j){
			int pi = e[i].size(), pj = e[j].size();
			int k = min(pi, pj);
			{
				map<int,int> v;
				for(auto u : e[i]) v[u]++;
				int upto = ((pi - k) / 2) * 2;
				for(auto p : v){
					int can = (p.second / 2) * 2;
					int now = min(can, upto); 
					ans += now / 2;
					upto -= now;
				}
			}
			{
				map<int,int> v;
				for(auto u : e[j]) v[u]++;
				int upto = ((pj - k) / 2) * 2;
				for(auto p : v){
					int can = (p.second / 2) * 2;
					int now = min(can, upto);
					ans += now / 2;
					upto -= now;
				}
			}
			ans += k;
		}
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Please create a solution in Cpp to 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;
const int maxn = 100100;
vector<int> e[maxn];
int main(){
	int n, m; cin >> n >> m;
	for(int i = 0; i < n; i++){
		int x; scanf("%d", &x);
		e[x % m].push_back(x);
	}
	int ans = e[0].size() / 2;
	for(int i = 1; i < m; i++){
		int j = m - i;
		if(i == j){
			ans += e[i].size() / 2;
		} else if(i < j){
			int pi = e[i].size(), pj = e[j].size();
			int k = min(pi, pj);
			{
				map<int,int> v;
				for(auto u : e[i]) v[u]++;
				int upto = ((pi - k) / 2) * 2;
				for(auto p : v){
					int can = (p.second / 2) * 2;
					int now = min(can, upto); 
					ans += now / 2;
					upto -= now;
				}
			}
			{
				map<int,int> v;
				for(auto u : e[j]) v[u]++;
				int upto = ((pj - k) / 2) * 2;
				for(auto p : v){
					int can = (p.second / 2) * 2;
					int now = min(can, upto);
					ans += now / 2;
					upto -= now;
				}
			}
			ans += k;
		}
	}
	cout << ans << endl;
	return 0;
}

```