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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <unordered_map>

using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n, m;
	cin>>n>>m;
	unordered_map<int, int> mp[100001];
	for(int i=0; i<n; i++){
		int x; cin>>x;
		mp[x%m][x]++;
	}
	int ans=0;
	int s=0;
	for(auto p:mp[0]){
		s+=p.second;
	}
	ans+=(s/2);
	if(m%2==0){
		s=0;
		for(auto p:mp[m/2]){
			s+=p.second;
		}
		ans+=(s/2);
	}
	for(int i=1; i<(m+1)/2; i++){
		int s1=0, s2=0, m1=0, m2=0;
		for(auto p:mp[i]){
			s1+=p.second;
			m1+=(p.second/2);
		}
		for(auto p:mp[m-i]){
			s2+=p.second;
			m2+=(p.second/2);
		}
		if(s1==s2){
			ans+=s1;
		}else if(s1<s2){
			if(2*m2>=s2-s1-1){
				ans+=((s1+s2)/2);
			}else{
				ans+=(s1+m2);
			}
		}else{
			if(2*m1>=s1-s2-1){
				ans+=((s1+s2)/2);
			}else{
				ans+=(s2+m1);
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}