#include<bits/stdc++.h>
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EPS (1e-10)
#define rep(i,n)for(ll i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int, int>P;

set<ll>val[100005];
ll cnt[100005], valc[100005];

int main() {
	ll n, m; scanf("%lld%lld", &n, &m);
	rep(i, n) {
		ll x; scanf("%lld", &x);
		val[x%m].insert(x); cnt[x%m]++; valc[x]++;
	}
	ll sum = cnt[0] / 2;
	if (m % 2 == 0)sum += cnt[m / 2] / 2;
	for (ll i = 1; i + i < m; i++) {
		ll c1 = 0, c2 = 0;
		for (ll j : val[i])c1 += valc[j] / 2;
		for (ll j : val[m - i])c2 += valc[j] / 2;
		ll Max = 0;
		for (ll j = 0; j <= c1; j++)for (ll k = 0; k <= c2; k++) {
			Max = max(Max, min(cnt[i] - j * 2, cnt[m - i] - k * 2) + j + k);
		}
		sum += Max;
	}
	printf("%lld\n", sum);
}
### Prompt

Generate a CPP solution 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>
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EPS (1e-10)
#define rep(i,n)for(ll i=0;i<n;i++)
using namespace std;
typedef long long ll;
typedef pair<int, int>P;

set<ll>val[100005];
ll cnt[100005], valc[100005];

int main() {
	ll n, m; scanf("%lld%lld", &n, &m);
	rep(i, n) {
		ll x; scanf("%lld", &x);
		val[x%m].insert(x); cnt[x%m]++; valc[x]++;
	}
	ll sum = cnt[0] / 2;
	if (m % 2 == 0)sum += cnt[m / 2] / 2;
	for (ll i = 1; i + i < m; i++) {
		ll c1 = 0, c2 = 0;
		for (ll j : val[i])c1 += valc[j] / 2;
		for (ll j : val[m - i])c2 += valc[j] / 2;
		ll Max = 0;
		for (ll j = 0; j <= c1; j++)for (ll k = 0; k <= c2; k++) {
			Max = max(Max, min(cnt[i] - j * 2, cnt[m - i] - k * 2) + j + k);
		}
		sum += Max;
	}
	printf("%lld\n", sum);
}
```