#include<bits/stdc++.h>
#define ll int64_t
#define read "input.txt","r",stdin
//#define write "output.txt","w",stdout
//#define m 1000000007
//#define limit 100005
using namespace std;

void sol(){
  ll n,x,a=0,sum=0,step=INT_MAX,cnt=0,ck=1;
  cin>>n>>x;
  ll ar[n+1],p[n+1];
  for(ll i=1;i<=n;i++){
    p[i]=x;
  }
  //for(ll i=1;i<=n;i++ )cout<<p[i];
  for(ll i=1;i<=n;i++){
    cin>>ar[i];
    sum+=ar[i];
  }
  while(ck){
        //cout<<ck<<endl<<sum<<endl;
    for(ll i=1;i<=n;i++){
        a=ar[i];
        if(a%x==0){
            sum+=(a/x)*p[i];
            a/=x;
            ar[i]=a;
            p[i]*=x;
        }
        else{
            ck=0;
            break;
        }
    }
  }
  cout<<sum<<endl;

}
int main(){
//    freopen(read);
     ios::sync_with_stdio(0);cin.tie(0);
     ll t=1;
     cin>>t;
     while(t--)
         sol();
     return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>
#define ll int64_t
#define read "input.txt","r",stdin
//#define write "output.txt","w",stdout
//#define m 1000000007
//#define limit 100005
using namespace std;

void sol(){
  ll n,x,a=0,sum=0,step=INT_MAX,cnt=0,ck=1;
  cin>>n>>x;
  ll ar[n+1],p[n+1];
  for(ll i=1;i<=n;i++){
    p[i]=x;
  }
  //for(ll i=1;i<=n;i++ )cout<<p[i];
  for(ll i=1;i<=n;i++){
    cin>>ar[i];
    sum+=ar[i];
  }
  while(ck){
        //cout<<ck<<endl<<sum<<endl;
    for(ll i=1;i<=n;i++){
        a=ar[i];
        if(a%x==0){
            sum+=(a/x)*p[i];
            a/=x;
            ar[i]=a;
            p[i]*=x;
        }
        else{
            ck=0;
            break;
        }
    }
  }
  cout<<sum<<endl;

}
int main(){
//    freopen(read);
     ios::sync_with_stdio(0);cin.tie(0);
     ll t=1;
     cin>>t;
     while(t--)
         sol();
     return 0;
}

```