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 <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <climits>
#include <ctime>
#include <cassert>
using namespace std;

#define rep(i,a,n) for(int i=a; i<n; i++)
#define repq(i,a,n) for(int i=a; i<=n; i++)
#define repr(i,a,n) for(int i=a; i>=n; i--)
#define pb(a) push_back(a)
#define fr first
#define sc second
#define INF 2000000000
#define int long long int

#define X real()
#define Y imag()
#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
#define EQV(a,b) ( EQ((a).X, (b).X) && EQ((a).Y, (b).Y) )
#define LE(n, m) ((n) < (m) + EPS)
#define LEQ(n, m) ((n) <= (m) + EPS)
#define GE(n, m) ((n) + EPS > (m))
#define GEQ(n, m) ((n) + EPS >= (m))

typedef vector<int> VI;
typedef vector<VI> MAT;
typedef pair<int, int> pii;
typedef long long ll;

typedef complex<double> P;
typedef pair<P, P> L;
typedef pair<P, double> C;

int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int const MOD = 1000000007;

namespace std {
    bool operator<(const P& a, const P& b) {
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

signed main() {
    int n, m, inp; cin >> n >> m;
    int x[100010] = {};
    int num[100010] = {};
    rep(i,0,n) {
        cin >> inp;
        x[inp % m]++;
        num[inp]++;
    }

    int ans = 0;
    rep(i,0,m) {
        int t;
        if(i != 0 && i != m-i) t = min(x[i], x[m-i]);
        else t = x[i] / 2;
        ans += t;
        if(i != 0 && i != m-i) {
            x[i] -= t; x[m-i] -= t;
        }
        else {
            x[i] -= t * 2;
        }
        // cout << ans << endl;
    }

    rep(i,0,m) {
        // printf("x[%lld] : %lld\n", i, x[i]);
        for(int j=i; j<=100000; j += m) {
            if(x[i] < 2) break;
            if(num[j] >= 2) {
                int temp = min(x[i], num[j]);
                ans += temp / 2;
                x[i] -= temp / 2 * 2;
            }
        }
    }
    cout << ans << endl;
    return 0;
}