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>
#define all(vec) vec.begin(),vec.end()
using namespace std;
typedef long long ll;
typedef pair<int,pair<int,int>> P;
const ll MOD=1000000007;
const ll INF=1000000010;
const ll LINF=4000000000000000010;
int dx[5]={0,1,0,-1,0};
int dy[5]={1,0,-1,0,0};
int c[100010];
vector<int> v[100010];
int main(){
    int n,m;cin>>n>>m;
    int a[100010];
    for(int i=0;i<n;i++){
        cin>>a[i];
        c[a[i]%m]++;
        v[a[i]%m].push_back(a[i]);
    }
    int ans=0;
    ans+=c[0]/2;
    for(int i=1;i<=m/2;i++){
        if(m%2==0&&i==m/2){
            ans+=c[i]/2;
        }else{
            ans+=min(c[i],c[m-i]);
            int s=abs(c[i]-c[m-i]);
            int cc=0;
            if(c[i]>c[m-i]){
                sort(all(v[i]));
                for(int j=1;j<v[i].size();j++){
                    if(v[i][j]==v[i][j-1]){
                        cc++;
                        j++;
                    }
                }
            }else{
                sort(all(v[m-i]));
                for(int j=1;j<v[m-i].size();j++){
                    if(v[m-i][j]==v[m-i][j-1]){
                        cc++;
                        j++;
                    }
                }
            }
            while(s>=2&&cc--){
                ans++;
                s-=2;
            }
        }
    }
    cout<<ans<<endl;
}
