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

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

int cnt[100010];

int main(int argc, char const* argv[])
{
  int n, m; cin >> n >> m;
  vector<int> x(n); cin >> x;
  rep(i, n){
    cnt[x[i]]++;
  }
  int res = 0;
  for(int i = 1; i <= m / 2; i++){
    if(i == m / 2 && m % 2 == 0)continue;
    int tmp = 0;
    queue<int> l1, l2, r1, r2;
    int L = 0, R = 0;
    for(int j = i; j <= 100000; j += m){
      if(cnt[j] & 1){
        l1.push(cnt[j]);
        tmp += cnt[j] / 2;
        L += cnt[j];
      }else{
        if(cnt[j] > 0){
          l2.push(cnt[j]);
          tmp += cnt[j] / 2;
          L += cnt[j];
        }
      }
    }
    for(int j = m - i; j <= 100000; j += m){
      if(cnt[j] & 1){
        r1.push(cnt[j]);
        tmp += cnt[j] / 2;
        R += cnt[j];
      }else{
        if(cnt[j] > 0){
          r2.push(cnt[j]);
          tmp += cnt[j] / 2;
          R += cnt[j];
        }
      }
    }
    int ans = tmp;
    for(int j = 1; j <= min(L, R); j++){
      tmp++;
      if(sz(l1)){
        int p = l1.front(); l1.pop();
        if(p > 1)l2.push(p - 1);
      }else{
        int p = l2.front(); l2.pop();
        tmp--;
        l1.push(p - 1);
      }
      if(sz(r1)){
        int p = r1.front(); r1.pop();
        if(p > 1)r2.push(p - 1);
      }else{
        int p = r2.front(); r2.pop();
        tmp--;
        r1.push(p - 1);
      }
      ans = max(ans, tmp);
    }
    res += ans;
  }
  int c = 0;
  for(int i = 0; i <= 100000; i += m){
    res += cnt[i] / 2;
    c += cnt[i] & 1;
  }
  res += c / 2;
  c = 0;
  if(m % 2 == 0){
    for(int i = m / 2; i <= 100000; i += m){
      res += cnt[i] / 2;
      c += cnt[i] & 1;
    }
    res += c / 2;
  }
  cout << res << endl;
  return 0;
}
