#include<bits/stdc++.h>
#define int long long
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
#define puts(x) cout << x << "\n"
using namespace std;
set<int>s;
int p[123456], n, m, a[123456], ans;
signed main() {
	cin >> n >> m;
	while (n--) {
		int x; cin >> x;
		if (s.find(x) != s.end()) {
			s.erase(x);
			p[x % m] += 2;
		}
		else s.insert(x);
	}
	for (int i : s)a[i % m]++;
	for1(i, (m + 1) / 2 - 1) {
		while (p[i] && a[i] + 2 <= a[m - i]) { p[i] -= 2; a[i] += 2; }
		while (p[m - i] && a[m - i] + 2 <= a[i]) { p[m - i] -= 2; a[m - i] += 2; }
		ans += p[i] / 2 + p[m - i] / 2 + min(a[i], a[m - i]);
	}
	ans += p[0] / 2 + a[0] / 2;
	if (m % 2 == 0)ans += p[m / 2] / 2 +a[m / 2] / 2;
	puts(ans);
}
### 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<bits/stdc++.h>
#define int long long
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
#define puts(x) cout << x << "\n"
using namespace std;
set<int>s;
int p[123456], n, m, a[123456], ans;
signed main() {
	cin >> n >> m;
	while (n--) {
		int x; cin >> x;
		if (s.find(x) != s.end()) {
			s.erase(x);
			p[x % m] += 2;
		}
		else s.insert(x);
	}
	for (int i : s)a[i % m]++;
	for1(i, (m + 1) / 2 - 1) {
		while (p[i] && a[i] + 2 <= a[m - i]) { p[i] -= 2; a[i] += 2; }
		while (p[m - i] && a[m - i] + 2 <= a[i]) { p[m - i] -= 2; a[m - i] += 2; }
		ans += p[i] / 2 + p[m - i] / 2 + min(a[i], a[m - i]);
	}
	ans += p[0] / 2 + a[0] / 2;
	if (m % 2 == 0)ans += p[m / 2] / 2 +a[m / 2] / 2;
	puts(ans);
}
```