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
/*
ID: y1197771
PROG: test
LANG: C++
*/
#include<bits/stdc++.h>
#define pb push_back
#define FOR(i, n) for (int i = 0; i < (int)n; ++i)
#define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
typedef long long ll;
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn], c[maxn];
int n, m;
void solve() {
    cin >> n >> m;
    int x, y;
    for (int i = 0; i < n; i++) {
        cin >> x;
        a[x]++;
        b[x % m]++;
    }
    for (int i = 1; i <= 1e5; i++) {
        c[i % m] += a[i] / 2;
    }
    ll res = 0;
    res += b[0] / 2;
    for (int i = 1; i <= m / 2; i++) {
        if(i == m - i) {
            res += b[i] / 2;
        } else {
            int mi = min(b[i], b[m - i]);
            int t = m - i;
            if(mi != b[i]) t = i;
            res += mi;
            int left = b[t] - mi;
            res += min(left / 2, c[t]);
        }
    }
    cout << res << endl;
}
int main() {
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    solve();
    return 0;
}
