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 all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

vector<int> XS[100010];

int main() {
	int N, M, X;
	cin >> N >> M;
	for (int i = 0;i < N;i++) {
		cin >> X;
		XS[X % M].push_back(X);
	}
	int ans = XS[0].size() / 2, t, cnt, cmax;
	if (M % 2 == 0)ans += XS[M / 2].size() / 2;
	for (int i = 1;i < (M + 1) / 2;i++) {
		cmax = 0, cnt = 1;
		if (XS[M - i].size() > XS[i].size()) {
			t = XS[M - i].size() - XS[i].size();
			sort(all(XS[M - i]));
			for (int j = 1;j < XS[M - i].size();j++) {
				if (XS[M - i][j] == XS[M - i][j - 1])cnt++;
				else cmax += cnt / 2, cnt = 1;
			}
			cmax += cnt / 2;
			ans += XS[i].size() + min(t / 2, cmax);
		}
		else {
			t = XS[i].size() - XS[M - i].size();
			sort(all(XS[i]));
			for (int j = 1;j < XS[i].size();j++) {
				if (XS[i][j] == XS[i][j - 1])cnt++;
				else cmax += cnt / 2, cnt = 1;
			}
			cmax += cnt / 2;
			ans += XS[M - i].size() + min(t / 2, cmax);
		}
	}
	cout << ans << endl;
}