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 sz(x) ((int) (x).size())
#define forn(i,n) for (int i = 0; i < int(n); ++i)
typedef long long ll;
typedef long long i64;
typedef long double ld;
const int inf = int(1e9) + int(1e5);
const ll infl = ll(2e18) + ll(1e10);

const int maxn = 100100;
vector<int> v[maxn];

int calc(vector<int> v) {
    int res = 0;
    forn (i, sz(v) - 1) {
        if (v[i] == v[i + 1]) {
            ++res;
            ++i;
        }
    }
    return res;
}

int main() {
    #ifdef LOCAL
    assert(freopen("d.in", "r", stdin));
    #else
    #endif
    int n, m;
    cin >> n >> m;
    forn (i, n) {
        int x;
        cin >> x;
        v[x % m].push_back(x);
    }
    forn (i, m)
        sort(v[i].begin(), v[i].end());
    int res = 0;
    for (int i = 0; i * 2 <= m; ++i) {
        if (i * 2 == m || i == 0) {
            res += sz(v[i]) / 2;
            continue;
        }
        int a = calc(v[i]);
        int b = calc(v[m - i]);
        int sa = sz(v[i]);
        int sb = sz(v[m - i]);
        int cur = min(sa, sb);
        sa -= cur, sb -= cur;
        cur += min(sa / 2, a);
        cur += min(sb / 2, b);
        res += cur;
    }
    cout << res << '\n';
}
