#include <iostream>
#include <vector>
#include <algorithm>
#include <map>

using namespace std;
vector<int> v[100010];
map<int,int> mp;
int main(){
	int i,j,n,m,x;
	long long ans=0;
	cin >> n >> m;
	for(i=0;i<n;i++){
		cin >> x;
		v[x%m].push_back(x);
		mp[x]++;
	}
	
	ans += v[0].size()/2;
	for(i=1;i<=(m-1)/2;i++){
		int cnt = 0,a = v[i].size(),b = v[m-i].size();
		ans += min(a,b);
		if(a>b){
			sort(v[i].begin(),v[i].end());
			v[i].erase(unique(v[i].begin(),v[i].end()),v[i].end());
			for(j=0;j<v[i].size();j++){
				cnt += mp[v[i][j]]/2;
			}
			
		}else{
			sort(v[m-i].begin(),v[m-i].end());
			v[m-i].erase(unique(v[m-i].begin(),v[m-i].end()),v[m-i].end());
				for(j=0;j<v[m-i].size();j++){
					cnt += mp[v[m-i][j]]/2;
				}
			}
		ans += min(max(a-b,b-a)/2,cnt);
	}
	if(m%2==0){
		ans += v[m/2].size()/2;
	}
	cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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 <vector>
#include <algorithm>
#include <map>

using namespace std;
vector<int> v[100010];
map<int,int> mp;
int main(){
	int i,j,n,m,x;
	long long ans=0;
	cin >> n >> m;
	for(i=0;i<n;i++){
		cin >> x;
		v[x%m].push_back(x);
		mp[x]++;
	}
	
	ans += v[0].size()/2;
	for(i=1;i<=(m-1)/2;i++){
		int cnt = 0,a = v[i].size(),b = v[m-i].size();
		ans += min(a,b);
		if(a>b){
			sort(v[i].begin(),v[i].end());
			v[i].erase(unique(v[i].begin(),v[i].end()),v[i].end());
			for(j=0;j<v[i].size();j++){
				cnt += mp[v[i][j]]/2;
			}
			
		}else{
			sort(v[m-i].begin(),v[m-i].end());
			v[m-i].erase(unique(v[m-i].begin(),v[m-i].end()),v[m-i].end());
				for(j=0;j<v[m-i].size();j++){
					cnt += mp[v[m-i][j]]/2;
				}
			}
		ans += min(max(a-b,b-a)/2,cnt);
	}
	if(m%2==0){
		ans += v[m/2].size()/2;
	}
	cout << ans << endl;
}

```