#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
int cnt[100010],mcnt[100010],n,m;
int main(){
    lol(i,100010)cnt[i]=0;
    cin>>n>>m;
    lol(i,n){
	int x;cin>>x;
	cnt[x]++;
	mcnt[x%m]++;
    }
    lol(i,100010){
	if(i<m)cnt[i]/=2;
	else cnt[i%m]+=cnt[i]/2;
    }
    int ans=mcnt[0]/2;
    for(int i=1;i<m;i++){
	int to=m-i;
	if(i>to)break;
	if(i==to){
	    ans+=mcnt[i]/2;
	    break;
	}
	if(mcnt[i]<mcnt[to]){
	    ans+=mcnt[i]+min((mcnt[to]-mcnt[i])/2,cnt[to]);
	}
	else{
	    ans+=mcnt[to]+min((mcnt[i]-mcnt[to])/2,cnt[i]);
	}
    }
    cout<<ans<<endl;
    //lol(i,m)cout<<cnt[i]<<" "<<mcnt[i]<<endl;
    return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
int cnt[100010],mcnt[100010],n,m;
int main(){
    lol(i,100010)cnt[i]=0;
    cin>>n>>m;
    lol(i,n){
	int x;cin>>x;
	cnt[x]++;
	mcnt[x%m]++;
    }
    lol(i,100010){
	if(i<m)cnt[i]/=2;
	else cnt[i%m]+=cnt[i]/2;
    }
    int ans=mcnt[0]/2;
    for(int i=1;i<m;i++){
	int to=m-i;
	if(i>to)break;
	if(i==to){
	    ans+=mcnt[i]/2;
	    break;
	}
	if(mcnt[i]<mcnt[to]){
	    ans+=mcnt[i]+min((mcnt[to]-mcnt[i])/2,cnt[to]);
	}
	else{
	    ans+=mcnt[to]+min((mcnt[i]-mcnt[to])/2,cnt[i]);
	}
    }
    cout<<ans<<endl;
    //lol(i,m)cout<<cnt[i]<<" "<<mcnt[i]<<endl;
    return 0;
}

```