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<string>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<math.h>
#include<time.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)

typedef long long i64;
typedef pair<i64,i64> P;


int n, m, a[101010], d[101010], b[101010];

int main(){
	cin >> n >> m;
	for(int i = 0; i < n; i++){
		cin >> a[i];
		d[a[i]%m]++;
		b[a[i]]++;
	}
	sort(a,a+n);
	a[n] = INF;
	
	int ans = 0;
	ans += d[0]/2;
	for(int i = 1; i < m/2+m%2; i++){
		ans += min(d[i], d[m-i]);
		int cnt1 = 0, cnt2 = 0, b1 = 0 , b2 = 0;
		for(int j = i; j < 101010; j+=m){
			cnt1 += b[j] / 2;
			b1 += b[j] % 2;
		}
		for(int j = m-i; j < 101010; j+=m){
			cnt2 += b[j] / 2;
			b2 += b[j] % 2;
		}
		ans += min(cnt1, (cnt1*2+b1 - min(d[i], d[m-i]))/2);
		ans += min(cnt2, (cnt2*2+b2 - min(d[i], d[m-i]))/2);
	}
	if(m%2 == 0){
		ans += d[m/2]/2;
	}
	
	cout << ans << endl;
	
	return 0;
}