#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <unordered_set>
#include <unordered_map>

using namespace std;

#define REP(i,n) for(int i=0; i<(int)(n); i++)
#define FOR(i,b,e) for (int i=(int)(b); i<(int)(e); i++)
#define ALL(x) (x).begin(), (x).end()

int n;
int m;
int cnt[100001];
int modcnt[100001];

// take r elements from the set of {x, x+m, x+2m,...}
void take(int x, int r) {
  for (int i = x; i < 100001; i += m) {
    if (cnt[i] % 2) {
      --cnt[i];
      if (!--r)
        return;
    }
  }

  for (int i = x; i < 100001; i += m) {
    int t = min(cnt[i], r);
    r -= t;
    cnt[i] -= t;
    if (!r)
      return;
  }
}

int main() {
  scanf(" %d %d", &n, &m);
  REP (i, n) {
    int tmp;
    scanf(" %d", &tmp);
    ++cnt[tmp];
    ++modcnt[tmp % m];
  }

  int ret = 0;
  REP (i, m) {
    int j = (m-i) % m;
    int z = (i == j) ? modcnt[i] / 2 : min(modcnt[i], modcnt[j]);
    if (z > 0) {
      modcnt[i] -= z;
      modcnt[j] -= z;
      ret += z;
      take(i, z);
      take(j, z);
    }
  }

  REP (i, 100001) {
    ret += cnt[i] / 2;
  }
  
  printf("%d\n", ret);
  
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <unordered_set>
#include <unordered_map>

using namespace std;

#define REP(i,n) for(int i=0; i<(int)(n); i++)
#define FOR(i,b,e) for (int i=(int)(b); i<(int)(e); i++)
#define ALL(x) (x).begin(), (x).end()

int n;
int m;
int cnt[100001];
int modcnt[100001];

// take r elements from the set of {x, x+m, x+2m,...}
void take(int x, int r) {
  for (int i = x; i < 100001; i += m) {
    if (cnt[i] % 2) {
      --cnt[i];
      if (!--r)
        return;
    }
  }

  for (int i = x; i < 100001; i += m) {
    int t = min(cnt[i], r);
    r -= t;
    cnt[i] -= t;
    if (!r)
      return;
  }
}

int main() {
  scanf(" %d %d", &n, &m);
  REP (i, n) {
    int tmp;
    scanf(" %d", &tmp);
    ++cnt[tmp];
    ++modcnt[tmp % m];
  }

  int ret = 0;
  REP (i, m) {
    int j = (m-i) % m;
    int z = (i == j) ? modcnt[i] / 2 : min(modcnt[i], modcnt[j]);
    if (z > 0) {
      modcnt[i] -= z;
      modcnt[j] -= z;
      ret += z;
      take(i, z);
      take(j, z);
    }
  }

  REP (i, 100001) {
    ret += cnt[i] / 2;
  }
  
  printf("%d\n", ret);
  
  return 0;
}

```