#include <bits/stdc++.h>
#define maxn 200009
using namespace std;
vector<int>G[maxn];
int solve(int x,int y){
    int cnt=0;
    int n=G[x].size();
    int cur=0;
    while(cur<n){
        int L=cur;
        while(cur+1<n&&G[x][cur]==G[x][cur+1])
            cur++;
        int R=cur;
        cnt+=(R-L+1)/2;
        cur++;
    }
    return min(y/2,cnt);
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        G[x%m].push_back(x);
    }
    for(int i=0;i<m;i++)
        sort(G[i].begin(),G[i].end());


    
    int ans=0;
    for(int i=0;i<=m/2;i++){
        int x=i;
        int y=(m-i)%m;
        int sz1=G[x].size();
        int sz2=G[y].size();
        if(x==y){
            ans+=sz1/2;
        }
        else if(y>x){
            if(sz1>sz2){
                ans+=solve(x,sz1-sz2);
            }
            else if(sz2>sz1){
                ans+=solve(y,sz2-sz1);
            }
            ans+=min(sz1,sz2);
        }
    }
    cout<<ans<<endl;
    //system("pause");
    return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <bits/stdc++.h>
#define maxn 200009
using namespace std;
vector<int>G[maxn];
int solve(int x,int y){
    int cnt=0;
    int n=G[x].size();
    int cur=0;
    while(cur<n){
        int L=cur;
        while(cur+1<n&&G[x][cur]==G[x][cur+1])
            cur++;
        int R=cur;
        cnt+=(R-L+1)/2;
        cur++;
    }
    return min(y/2,cnt);
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        G[x%m].push_back(x);
    }
    for(int i=0;i<m;i++)
        sort(G[i].begin(),G[i].end());


    
    int ans=0;
    for(int i=0;i<=m/2;i++){
        int x=i;
        int y=(m-i)%m;
        int sz1=G[x].size();
        int sz2=G[y].size();
        if(x==y){
            ans+=sz1/2;
        }
        else if(y>x){
            if(sz1>sz2){
                ans+=solve(x,sz1-sz2);
            }
            else if(sz2>sz1){
                ans+=solve(y,sz2-sz1);
            }
            ans+=min(sz1,sz2);
        }
    }
    cout<<ans<<endl;
    //system("pause");
    return 0;
}
```