#include <algorithm>
#include <complex>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n) for (int i = (n)-1; i >= 0; i--)
using namespace std;
typedef long long int lli;
typedef pair<lli, lli> P;
lli MOD = 1000000007;
int main()
{
    lli n, m;
    cin >> n >> m;
    lli a;
    lli mod[100005] = {};
    lli x[100005] = {};
    rep(i, n)
    {
        cin >> a;
        mod[a % m]++;
        x[a]++;
    }
    lli ans = 0;
    rep(i, m)
    {
        if (i == 0) {
            ans += mod[i] / 2;
            mod[i] = mod[i] % 2;
        } else if (m % 2 == 0 && i == m / 2) {
            ans += mod[i] / 2;
            mod[i] = mod[i] % 2;
        } else if (i != 0 && !(m % 2 == 0 && i == m / 2)) {
            ans += min(mod[i], mod[(m - i) % m]);
            lli del = min(mod[i], mod[(m - i) % m]);
            mod[i]
                -= del;

            mod[(m - i) % m] -= del;
        }
    }
    rep(i, 100005)
    {
        if (x[i] < 2)
            continue;
        else {
            int l = i % m;
            int ma = mod[l] / 2;
            int use = x[i] / 2;
            if (ma >= use) {
                mod[l] -= use * 2;
                ans += use;
            } else {
                mod[l] = 0;
                ans += ma;
            }
        }
    }
    cout << ans << endl;
}

### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include <algorithm>
#include <complex>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n) for (int i = (n)-1; i >= 0; i--)
using namespace std;
typedef long long int lli;
typedef pair<lli, lli> P;
lli MOD = 1000000007;
int main()
{
    lli n, m;
    cin >> n >> m;
    lli a;
    lli mod[100005] = {};
    lli x[100005] = {};
    rep(i, n)
    {
        cin >> a;
        mod[a % m]++;
        x[a]++;
    }
    lli ans = 0;
    rep(i, m)
    {
        if (i == 0) {
            ans += mod[i] / 2;
            mod[i] = mod[i] % 2;
        } else if (m % 2 == 0 && i == m / 2) {
            ans += mod[i] / 2;
            mod[i] = mod[i] % 2;
        } else if (i != 0 && !(m % 2 == 0 && i == m / 2)) {
            ans += min(mod[i], mod[(m - i) % m]);
            lli del = min(mod[i], mod[(m - i) % m]);
            mod[i]
                -= del;

            mod[(m - i) % m] -= del;
        }
    }
    rep(i, 100005)
    {
        if (x[i] < 2)
            continue;
        else {
            int l = i % m;
            int ma = mod[l] / 2;
            int use = x[i] / 2;
            if (ma >= use) {
                mod[l] -= use * 2;
                ans += use;
            } else {
                mod[l] = 0;
                ans += ma;
            }
        }
    }
    cout << ans << endl;
}

```