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
#include <bits/stdc++.h>
#define LL long long
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
using namespace std;

int n,m,x,y,a[100100],ans,cnt[100100],cntPair[100100];
vector<int> v[100100];

int main(){
	scanf("%d %d",&n,&m);
	for(int i=0;i<n;i++) scanf("%d",&a[i]);
	sort(a,a+n);
	for(int i=0;i<n;i++){
		cnt[a[i]%m]++;
		v[a[i]%m].pb(a[i]);
	}
	ans+=cnt[0]/2;
	for(int i=1;i<m;i++){
		for(int j=0;j<v[i].size();j++){
			if(j!=v[i].size()-1 && v[i][j]==v[i][j+1]){
				cntPair[i]++;
				j++;
			}
		}
		//for(int j=0;j<v[i].size();j++) cout<<v[i][j]<<" ";cout<<endl;
		if(i*2==m) x = cnt[i]/2;
		else x = min(cnt[i],cnt[m-i]);
		cnt[i]-=x;
		cnt[m-i]-=x;
		ans+=x;
		ans+=min(cntPair[i],cnt[i]/2);
		//cout<<ans<<" "<<x<<" "<<cnt[i]<<" "<<cntPair[i]<<endl;
	}
	cout<<ans<<endl;
	
	return 0;
}