#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
    int n, m;
    cin >> n >> m;
    map<int, int> um;
    for(int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        um[x]++;
    }
    vector<int> modm(m);
    vector<int> mul(m);
    auto itr = um.begin();
    while(itr != um.end())
    {
        int v = itr->first;
        int count = itr->second;
        modm[v % m] += count;
        mul[v % m] += count / 2;
        itr++;
    }

    int ans = modm[0] / 2;
    for(int i = 1; i <= m/2; i++)
    {
        if(i == m/2 && m%2 == 0)
        {
            ans += modm[i] / 2;
            continue;
        }
        int pos;
        if(modm[i] > modm[m-i]) pos = i;
        else pos = m-i;
        int h = max(modm[i], modm[m-i]);
        int l = min(modm[i], modm[m-i]);
        ans += l;
        int more = h - l;
        ans += min(more / 2, mul[pos]);
    }
    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 <bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
    int n, m;
    cin >> n >> m;
    map<int, int> um;
    for(int i = 0; i < n; i++)
    {
        int x;
        cin >> x;
        um[x]++;
    }
    vector<int> modm(m);
    vector<int> mul(m);
    auto itr = um.begin();
    while(itr != um.end())
    {
        int v = itr->first;
        int count = itr->second;
        modm[v % m] += count;
        mul[v % m] += count / 2;
        itr++;
    }

    int ans = modm[0] / 2;
    for(int i = 1; i <= m/2; i++)
    {
        if(i == m/2 && m%2 == 0)
        {
            ans += modm[i] / 2;
            continue;
        }
        int pos;
        if(modm[i] > modm[m-i]) pos = i;
        else pos = m-i;
        int h = max(modm[i], modm[m-i]);
        int l = min(modm[i], modm[m-i]);
        ans += l;
        int more = h - l;
        ans += min(more / 2, mul[pos]);
    }
    cout << ans << endl;
}

```