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 <iostream>
#include <iomanip>
#include <ios> 
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <bitset>
#include <map>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define sz(c) ((int)(c).size())

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
const int MAX=1e5+10;

int main(){
	int N,M;
	cin>>N>>M;
	int ct[MAX];
	rep1(i,MAX-1)ct[i]=0;
	rep(_,N){
		int x;
		cin>>x;
		ct[x]++;
	}
	int card[MAX];
	int pair[MAX];
	rep(i,M){
		card[i]=0;
		pair[i]=0;
	}
	rep1(i,MAX-1){
		card[i%M]+=ct[i];
		pair[i%M]+=ct[i]/2;
	}
	int ans=card[0]/2;
	if(M%2==0){
		ans+=card[M/2]/2;
		rep1(i,M/2-1){
			int ac=card[i], ap=pair[i];
			int bc=card[M-i], bp=pair[M-i];
			int d1=ac-(bc-2*bp);
			int d2=(ac-2*ap)-bc;
			if(d1>0 && d2>0){
				ans+=(ac+bc-min(abs(d1), abs(d2)))/2;
			}
			else if(d1<0 && d2<0){
				ans+=(ac+bc-min(abs(d1), abs(d2)))/2;
			}
			else{
				ans+=(ac+bc-(ac+bc)%2)/2;
			}
		}
	}
	if(M%2==1){
		rep1(i,M/2){
			int ac=card[i], ap=pair[i];
			int bc=card[M-i], bp=pair[M-i];
			int d1=ac-(bc-2*bp);
			int d2=(ac-2*ap)-bc;
			if(d1>0 && d2>0){
				ans+=(ac+bc-min(abs(d1), abs(d2)))/2;
			}
			else if(d1<0 && d2<0){
				ans+=(ac+bc-min(abs(d1), abs(d2)))/2;
			}
			else{
				ans+=(ac+bc-(ac+bc)%2)/2;
			}
		}
	}
	cout<<ans<<endl;
}