#include <iostream>
#include <map>
#include <vector>
using namespace std;

int main() {
    int N, M;
    cin >> N >> M;

    vector<int> p_cnt(M, 0),
                o_cnt(M, 0);
    map<int, int> mp;
    for (int i = 0; i < N; i++) {
        int X;
        cin >> X;
        mp[X] ^= 1;
        if (mp[X] == 0) {
            p_cnt[X % M]++;
        }
    }
    for (auto p : mp) o_cnt[p.first % M] += p.second;

    int ans = 0;
    for (int i1 = 0; 2 * i1 <= M; i1++) {
        int i2 = (M - i1) % M;
        if (i1 == i2) {
            ans += (2 * p_cnt[i1] + o_cnt[i1]) / 2;
        } else {
            int num1 = 2 * p_cnt[i1] + o_cnt[i1],
                num2 = 2 * p_cnt[i2] + o_cnt[i2];
            if (num1 < num2) {
                swap(num1, num2);
                swap(p_cnt[i1], p_cnt[i2]);
                swap(o_cnt[i1], o_cnt[i2]);
            }
            ans += min(num1, num2);
            ans += min(p_cnt[i1], (num1 - num2) / 2);
        }
    }
    cout << ans << endl;

    return 0;
}
### Prompt

Your challenge is to write 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 <map>
#include <vector>
using namespace std;

int main() {
    int N, M;
    cin >> N >> M;

    vector<int> p_cnt(M, 0),
                o_cnt(M, 0);
    map<int, int> mp;
    for (int i = 0; i < N; i++) {
        int X;
        cin >> X;
        mp[X] ^= 1;
        if (mp[X] == 0) {
            p_cnt[X % M]++;
        }
    }
    for (auto p : mp) o_cnt[p.first % M] += p.second;

    int ans = 0;
    for (int i1 = 0; 2 * i1 <= M; i1++) {
        int i2 = (M - i1) % M;
        if (i1 == i2) {
            ans += (2 * p_cnt[i1] + o_cnt[i1]) / 2;
        } else {
            int num1 = 2 * p_cnt[i1] + o_cnt[i1],
                num2 = 2 * p_cnt[i2] + o_cnt[i2];
            if (num1 < num2) {
                swap(num1, num2);
                swap(p_cnt[i1], p_cnt[i2]);
                swap(o_cnt[i1], o_cnt[i2]);
            }
            ans += min(num1, num2);
            ans += min(p_cnt[i1], (num1 - num2) / 2);
        }
    }
    cout << ans << endl;

    return 0;
}
```