#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long int ulli;
typedef vector<long long> vi;
typedef vector< pair<int,int>> vp;
typedef pair<int,int> pi;
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define f(i,a,b) for (long long i = a; i <= b; i++)
//vi a;for (auto &it : a) cin >> it;
vi ar(200);
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) 
{ 
    return (a.second < b.second); 
}
int mod=100000;
void solve()
{
     ll n,x;
     cin>>n>>x;
     ll a[n],b[n];
     for(int i=0;i<n;i++){
         cin>>a[i];ll r=0;
         int y=a[i];
         while(y%x==0){
             r++;y/=x;
         }
         b[i]=r;
     }
     ll mn=b[0],pos=0;ll ans=0;
     for(int i=1;i<n;i++){
         if(mn>b[i]){
             mn=b[i];pos=i;
         }
     }
     for(int i=0;i<pos;i++){
         ll p=((mn+2)*a[i]);
        ans+=p;
     }
     for(int i=pos;i<n;i++){
        ll p=((mn+1)*a[i]);
        ans+=p;
     }
     cout<<ans<<endl;
return;
}
int main()
{
    int t;cin>>t;
    while(t--)
    {
         solve();
    }
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

You have given an array a of length n and an integer x to a brand new robot. What the robot does is the following: it iterates over the elements of the array, let the current element be q. If q is divisible by x, the robot adds x copies of the integer q/x to the end of the array, and moves on to the next element. Note that the newly added elements could be processed by the robot later. Otherwise, if q is not divisible by x, the robot shuts down.

Please determine the sum of all values of the array at the end of the process.

Input

The first input line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the length of the array and the value which is used by the robot.

The next line contains integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the initial values in the array.

It is guaranteed that the sum of values n over all test cases does not exceed 10^5.

Output

For each test case output one integer — the sum of all elements at the end of the process.

Example

Input


2
1 2
12
4 2
4 6 8 2


Output


36
44

Note

In the first test case the array initially consists of a single element [12], and x=2. After the robot processes the first element, the array becomes [12, 6, 6]. Then the robot processes the second element, and the array becomes [12, 6, 6, 3, 3]. After the robot processes the next element, the array becomes [12, 6, 6, 3, 3, 3, 3], and then the robot shuts down, since it encounters an element that is not divisible by x = 2. The sum of the elements in the resulting array is equal to 36.

In the second test case the array initially contains integers [4, 6, 8, 2], and x=2. The resulting array in this case looks like  [4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1].

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long int ulli;
typedef vector<long long> vi;
typedef vector< pair<int,int>> vp;
typedef pair<int,int> pi;
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define f(i,a,b) for (long long i = a; i <= b; i++)
//vi a;for (auto &it : a) cin >> it;
vi ar(200);
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) 
{ 
    return (a.second < b.second); 
}
int mod=100000;
void solve()
{
     ll n,x;
     cin>>n>>x;
     ll a[n],b[n];
     for(int i=0;i<n;i++){
         cin>>a[i];ll r=0;
         int y=a[i];
         while(y%x==0){
             r++;y/=x;
         }
         b[i]=r;
     }
     ll mn=b[0],pos=0;ll ans=0;
     for(int i=1;i<n;i++){
         if(mn>b[i]){
             mn=b[i];pos=i;
         }
     }
     for(int i=0;i<pos;i++){
         ll p=((mn+2)*a[i]);
        ans+=p;
     }
     for(int i=pos;i<n;i++){
        ll p=((mn+1)*a[i]);
        ans+=p;
     }
     cout<<ans<<endl;
return;
}
int main()
{
    int t;cin>>t;
    while(t--)
    {
         solve();
    }
	return 0;
}
```