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>
using namespace std;

#define FOR(i,l,r) for(int i = (l);i < (r);i++)
#define ALL(x) (x).begin(),(x).end()
template<typename T> bool chmax(T& a,const T& b){return a < b ? (a = b,true) : false;}
template<typename T> bool chmin(T& a,const T& b){return b < a ? (a = b,true) : false;}
typedef long long ll;

int N,M;
int cnt [100001];

int main()
{
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	cin >> N >> M;
	vector< vector<int> > A(M);
	vector<int> p(M);
	FOR(i,0,N){
		int X;
		cin >> X;
		if(++cnt [X] % 2 == 0){
			p [X % M]++;
		}
		A [X % M].push_back(X);
	}

	int ans = 0;
	for(int i = 0;i <= M / 2;i++){
		int j = (M - i) % M;
		if(i == j){
			ans += A [i].size() / 2;
		}
		else{
			int x = A [i].size(),y = A [j].size();
			if(x >= y){
				ans += min(x,y);
				ans += min((x - y) / 2,p [i]);
			}
			else{
				ans += min(x,y);
				ans += min((y - x) / 2,p [j]);
			}
		}
	}

	cout << ans << endl;

	return 0;
}
