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
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

void solve(ll test_num) {
  (void) test_num;
  const ll MAX = 111000;
  static vector<ll> by_mod[MAX];
  ll N, M;
  cin >> N >> M;
  for (ll i = 0, x; i < N; ++i) {
    cin >> x;
    by_mod[x % M].push_back(x);
  }
  ll res = 0;
  for (ll r = 0; r < M; ++r) {
    if (2 * r == M || r == 0) {
      res += by_mod[r].size() / 2;
    } else {
      const ll am = min(by_mod[r].size(), by_mod[M - r].size());
      if (r < M - r)
        res += am;
      sort(by_mod[r].begin(), by_mod[r].end());
      for (ll i = 1, cnt = 0; i < (ll) by_mod[r].size() && cnt + am + 2 <= by_mod[r].size(); ) {
        if (by_mod[r][i] == by_mod[r][i - 1]) {
          ++res;
          cnt += 2;
          i += 2;
        } else {
          ++i;
        }
      }
    }
  }
  cout << res << endl;
}

void init() {

}

int main() {
#ifdef AZN
  freopen("/home/azneye/Documents/Stuff/input.txt", "r", stdin);
#endif
  init();
  ll tests = 1;
  //cin >> tests;
  for (ll test = 1; test <= tests; ++test) {
    solve(test);
  }
  return 0;
}