#include <bits/stdc++.h>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long ll;

int n,m,r;
int q[100005],w[100005],z[100005];

int main(){
	scanf("%d%d",&n,&m);
	fore(i,0,n){
		int x;
		scanf("%d",&x);
		q[x]++;
		w[x%m]++;
	}
	fore(i,0,m){
		int j=(m-i)%m;
		if(i>j)break;
		if(i==j){
			r+=w[i]/2;
			z[i]=w[i]%2;
		}
		else {
			int k=min(w[i],w[j]);
			r+=k;z[i]=w[i]-k;z[j]=w[j]-k;
		}
	}
	fore(i,1,100005){
		int k=min(q[i],z[i%m])/2;
		r+=k;
		z[i%m]-=2*k;
	}
	printf("%d\n",r);
	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>
#ifdef DEMETRIO
#define deb(...) fprintf(stderr,__VA_ARGS__)
#define deb1(x) cerr << #x << " = " << x << endl
#else
#define deb(...) 0
#define deb1(x) 0
#endif
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
#define SZ(x) ((int)x.size())
using namespace std;
typedef long long ll;

int n,m,r;
int q[100005],w[100005],z[100005];

int main(){
	scanf("%d%d",&n,&m);
	fore(i,0,n){
		int x;
		scanf("%d",&x);
		q[x]++;
		w[x%m]++;
	}
	fore(i,0,m){
		int j=(m-i)%m;
		if(i>j)break;
		if(i==j){
			r+=w[i]/2;
			z[i]=w[i]%2;
		}
		else {
			int k=min(w[i],w[j]);
			r+=k;z[i]=w[i]-k;z[j]=w[j]-k;
		}
	}
	fore(i,1,100005){
		int k=min(q[i],z[i%m])/2;
		r+=k;
		z[i%m]-=2*k;
	}
	printf("%d\n",r);
	return 0;
}
```