#include <bits/stdc++.h>
#define sp ' '
#define nyan "(=^・ω・^=)"
#define mkp make_pair
#define intmax 2147483647
#define llmax 9223372036854775807
#define lP pair<ll,ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;
const int mod = 1000000007;
const int mod998 = 998244353;

int N, M, X, res;
vector<int>v[100000];

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; ++i) {
		cin >> X;
		v[X % M].push_back(X);
	}
	res += v[0].size() / 2;
	for (int i = 1; i < M; ++i) {
		if (i + i == M) {
			res += v[i].size() / 2;
		}
		else if (v[i].size() > v[M - i].size() || (v[i].size() == v[M - i].size() && i <= M - i)) {
			res += v[M - i].size();
			map<int, int>mp;
			for (int j : v[i]) {
				++mp[j];
			}
			int c = 0;
			for (auto it = mp.begin(); it != mp.end(); ++it) {
				c += it->second / 2;
			}
			res += min(int(v[i].size() - v[M - i].size()) / 2, c);
		}
	}
	cout << res << endl;
	return 0;
}
### 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>
#define sp ' '
#define nyan "(=^・ω・^=)"
#define mkp make_pair
#define intmax 2147483647
#define llmax 9223372036854775807
#define lP pair<ll,ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;
const int mod = 1000000007;
const int mod998 = 998244353;

int N, M, X, res;
vector<int>v[100000];

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; ++i) {
		cin >> X;
		v[X % M].push_back(X);
	}
	res += v[0].size() / 2;
	for (int i = 1; i < M; ++i) {
		if (i + i == M) {
			res += v[i].size() / 2;
		}
		else if (v[i].size() > v[M - i].size() || (v[i].size() == v[M - i].size() && i <= M - i)) {
			res += v[M - i].size();
			map<int, int>mp;
			for (int j : v[i]) {
				++mp[j];
			}
			int c = 0;
			for (auto it = mp.begin(); it != mp.end(); ++it) {
				c += it->second / 2;
			}
			res += min(int(v[i].size() - v[M - i].size()) / 2, c);
		}
	}
	cout << res << endl;
	return 0;
}
```