#include<bits/stdc++.h>
using namespace std;

int N,M;
int X[100010];
int cnt[100010];
int p[100010];
int num[100010];

int main()
{
    cin >> N >> M;
    for(int i = 0; i < N; i++)
    {
        int x;
        cin >> x;
        cnt[x]++;

    }
    for(int i = 1; i < 100010; i++)
    {
        p[i % M] += cnt[i] / 2;
        num[i % M] += cnt[i];
    }

    int ans = 0;
    ans += num[0] / 2;
    if(M % 2 == 0)ans += num[M / 2] / 2;
    for(int i = 1; i <= M / 2; i++)
    {
        if(M % 2 == 0 && M / 2 == i)continue;
        int x = num[i], y = num[M - i];
        bool flip = false;
        if(x < y)
        {
            swap(x, y);
            flip = true;
        }
        int d = x - y;
        ans += min(d / 2, p[flip ? M - i : i]);
        ans += y;
    }

    cout << ans << endl;
    return 0;

}

### Prompt

Create a solution in Cpp for 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;

int N,M;
int X[100010];
int cnt[100010];
int p[100010];
int num[100010];

int main()
{
    cin >> N >> M;
    for(int i = 0; i < N; i++)
    {
        int x;
        cin >> x;
        cnt[x]++;

    }
    for(int i = 1; i < 100010; i++)
    {
        p[i % M] += cnt[i] / 2;
        num[i % M] += cnt[i];
    }

    int ans = 0;
    ans += num[0] / 2;
    if(M % 2 == 0)ans += num[M / 2] / 2;
    for(int i = 1; i <= M / 2; i++)
    {
        if(M % 2 == 0 && M / 2 == i)continue;
        int x = num[i], y = num[M - i];
        bool flip = false;
        if(x < y)
        {
            swap(x, y);
            flip = true;
        }
        int d = x - y;
        ans += min(d / 2, p[flip ? M - i : i]);
        ans += y;
    }

    cout << ans << endl;
    return 0;

}

```