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;

#define int long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N,M;
int cnt[111111];
int odd[111111];

signed main(){
    cin>>N>>M;
    rep(i,N){
        int x;cin>>x;
        cnt[x]++;
    }

    rep(i,100000+1)if(cnt[i]&1)odd[i%M]++,cnt[i]--;
    int ans=0;
    for(int i=0;i<=M-i;i++){
        int a=i,b=(M-i)%M;
        if(a!=b){
            int t=min(odd[a],odd[b]);
            ans+=t;
            odd[a]-=t;
            odd[b]-=t;
        }
        else{
            ans+=odd[i]/2;
            odd[i]%=2;
        }
    }

    for(int i=M;i<=100000;i++)cnt[i%M]+=cnt[i];

    rep(i,M){
        int t=odd[i];
        if(t&1)t--;
        int m=min(t,cnt[(M-i)%M]);
        cnt[(M-i)%M]-=m;
        ans+=m;
    }
    rep(i,M)ans+=cnt[i]/2;
    cout<<ans<<endl;
    return 0;
}
