#include <bits/stdc++.h>
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priority_queue
#define P pair<ll,ll>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const ll LINF=1e18;
const ld PI=abs(acos(-1));
int n,m,a,lia[100010],lib[100010];

int main(){
	cin >> n >> m;
	REP(i,0,n) cin >> a,lia[a]++,lib[a%m]++;
	
	int s=lib[0]/2+(!(m%2)?lib[m/2]/2:0);
	for(int i=1;i*2<m;i++){
		int pos=min(lib[i], lib[m-i]);
		lib[i]-=pos;
		lib[m-i]-=pos;
		s+=pos;
	}
	
	REP(i,1,100010){
		if(!(i%m) or (i%m)*2==m) continue;
		int pos=min(lib[i%m]/2, lia[i]/2);
		s+=pos;
		lib[i%m]-=pos*2;
	}
	
	cout << s << 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 REP(i,a,b) for(int i=(a);i<(b);i++)
#define RREP(i,a,b) for(int i=(a);i>=(b);i--)
#define pq priority_queue
#define P pair<ll,ll>
#define P2 pair<int,P>
#define P3 pair<int,P2>
typedef long long ll; typedef long double ld;
using namespace std;
const int INF=1e9, MOD=1e9+7, around[]={0,1,1,-1,-1,0,-1,1,0,0};
const ll LINF=1e18;
const ld PI=abs(acos(-1));
int n,m,a,lia[100010],lib[100010];

int main(){
	cin >> n >> m;
	REP(i,0,n) cin >> a,lia[a]++,lib[a%m]++;
	
	int s=lib[0]/2+(!(m%2)?lib[m/2]/2:0);
	for(int i=1;i*2<m;i++){
		int pos=min(lib[i], lib[m-i]);
		lib[i]-=pos;
		lib[m-i]-=pos;
		s+=pos;
	}
	
	REP(i,1,100010){
		if(!(i%m) or (i%m)*2==m) continue;
		int pos=min(lib[i%m]/2, lia[i]/2);
		s+=pos;
		lib[i%m]-=pos*2;
	}
	
	cout << s << endl;
	return 0;
}

```