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;
typedef pair<int,int> P;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n,m;cin>>n>>m;
    map<int,int> mp;//(値, 個数)
    vector<set<int>> st(m);// st[i]:=余りiの値集合 
    vector<int> sum(m,0),ps(m,0);//sum[i],ps[i]:=余りiの個数,ペア数
    for(int i=0;i<n;i++){
        int x;cin>>x;
        mp[x]++;
        st[x%m].insert(x);
        sum[x%m]++;
    }
    int res=0;
    res += sum[0]/2;
    if(m%2==0){
        res += sum[m/2]/2;
        st[m/2].clear();
    }
    for(int i=1;i<m;i++){
        if(st[i].size()==0) continue;
        for(auto itr=st[i].begin();itr!=st[i].end();itr++){
            int x=*itr;
            ps[i] += mp[x]/2;
        }
    }
    for(int i=1;i<=m/2;i++){
        if(m%2==0 && i==m/2) continue;
        int j=m-i;
        if(sum[i]<sum[j]){
            swap(sum[i],sum[j]);
            swap(ps[i],ps[j]);
        }
        res += sum[j];
        res += min((sum[i]-sum[j])/2, ps[i]);
    }
    cout<<res<<endl;
}
