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 fo(i,a,b) for(int i=(a);i<(b);i++)
#define MOD 1000000007
#define MP make_pair
#define PB push_back
typedef long long ll;

int N, M, cnt[100100], g[100100], odd[100100], ans;
vector<int> v[100100], n[100100];

int main () {
        scanf("%d %d", &N, &M);
        fo(i, 0, N) {
                int x;
                scanf("%d", &x);
                cnt[x%M]++;
                v[x%M].PB(x);
        }
        fo(i, 0, M) {
                sort(v[i].begin(), v[i].end());
                fo(j, 0, (int) v[i].size()) {
                        if (!j || v[i][j-1] != v[i][j]) {
                                int x = j; while (x+1 < (int) v[i].size() && v[i][x+1] == v[i][j]) x++;
                                n[i].PB(x-j+1);
                                if (n[i].back()&1) odd[i]++, g[i] += x-j;
                                else g[i] += x-j+1;
                        }
                }
        }

        ans = cnt[0]/2;
        if (M%2 == 0) ans += cnt[M/2]/2;

        for (int i = 1; i < M-i; i++) {
                int a = i, b = M-i;
                if (odd[a] > odd[b]) swap(a, b);
                ans += odd[a], odd[b] -= odd[a], odd[a] = 0;
                while (odd[b] >= 2 && g[a] >= 2) odd[b] -= 2, g[a] -= 2, ans += 2;
                ans += g[a]/2 + g[b]/2;
        }

        printf("%d\n", ans);
        return 0;
}
