#include <iostream>
#include <vector>
#include <map>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
using namespace std;
int main() {
    // input
    int n, m; cin >> n >> m;
    map<int,int> xs;
    repeat (i,n) {
        int x; cin >> x;
        xs[x] += 1;
    }
    // compute
    vector<int> cnt(m), pr(m);
    for (auto it : xs) {
        int x, k; tie(x, k) = it;
        cnt[x % m] += k;
        pr [x % m] += k / 2 * 2;
    }
    int ans = 0;
    repeat (i,m) {
        int j = (m - i) % m;
        if (i == j) {
            ans += cnt[i] / 2;
        } else if (i < j) {
            int k = min(cnt[i], cnt[j]);
            ans += k;
            ans += min(cnt[i] - k, pr[i]) / 2;
            ans += min(cnt[j] - k, pr[j]) / 2;
        }
    }
    // output
    cout << ans << endl;
    return 0;
}

### Prompt

Generate 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 <iostream>
#include <vector>
#include <map>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
using namespace std;
int main() {
    // input
    int n, m; cin >> n >> m;
    map<int,int> xs;
    repeat (i,n) {
        int x; cin >> x;
        xs[x] += 1;
    }
    // compute
    vector<int> cnt(m), pr(m);
    for (auto it : xs) {
        int x, k; tie(x, k) = it;
        cnt[x % m] += k;
        pr [x % m] += k / 2 * 2;
    }
    int ans = 0;
    repeat (i,m) {
        int j = (m - i) % m;
        if (i == j) {
            ans += cnt[i] / 2;
        } else if (i < j) {
            int k = min(cnt[i], cnt[j]);
            ans += k;
            ans += min(cnt[i] - k, pr[i]) / 2;
            ans += min(cnt[j] - k, pr[j]) / 2;
        }
    }
    // output
    cout << ans << endl;
    return 0;
}

```