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 <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<int> xs(n);
    for (auto &x: xs)
        cin >> x;

    vector<vector<int>> bkt(m);
    for (auto &x: xs) {
        bkt[x % m].push_back(x);
    }
    for (auto &b: bkt) {
        sort(b.begin(), b.end());
    }

    int ans = 0;
    for (int i = 0; i < m; i++) {
        int j = (m - i) % m;

        if (i > j)
            continue;

        if (i == j) {
            ans += bkt[i].size() / 2;
            continue;
        }

        int p = 0;
        for (int k = 0; k + 1 < bkt[i].size(); k++) {
            if (bkt[i][k] == bkt[i][k + 1]) {
                p++;
                k++;
            }
        }
        int q = 0;
        for (int k = 0; k + 1 < bkt[j].size(); k++) {
            if (bkt[j][k] == bkt[j][k + 1]) {
                q++;
                k++;
            }
        }

        int is = bkt[i].size();
        int js = bkt[j].size();

        if (is > js) {
            swap(is, js);
            swap(p, q);
        }

        int sel = min((js - is) / 2, q);

        ans += sel;
        js -= sel * 2;
        ans += min(is, js);
    }

    cout << ans << endl;

    return 0;
}