#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define ll long long 
#define ff first
#define ss second
#define float long double

void solve()
{
    ll n,x,a,s=0;
    cin>>n>>x;
    vector<ll>v;
    for(ll i=0;i<n;i++)
   {
     cin>>a;
     v.push_back(a);
     
   }
    for(ll i=0;i<v.size();i++)
    {
        if(v[i]%x==0)
       { v.push_back(v[i]/x);
        if(i<n) s+=v[i]+v[i];
        else
        s+=((pow(x,i/n))*v[i]);
       }
        else
        {
            if(i<n)
            {
                for(ll j=i;j<n;j++)
                s+=v[j];
                
            }
            
            break;
        }
    }
    cout<<s<<endl;
    
   
         
     }

int main()
{
          ios_base::sync_with_stdio(false);
          cin.tie(0);
          cout.tie();
          int t;
          cin>>t;
          //t=1;
          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
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define ll long long 
#define ff first
#define ss second
#define float long double

void solve()
{
    ll n,x,a,s=0;
    cin>>n>>x;
    vector<ll>v;
    for(ll i=0;i<n;i++)
   {
     cin>>a;
     v.push_back(a);
     
   }
    for(ll i=0;i<v.size();i++)
    {
        if(v[i]%x==0)
       { v.push_back(v[i]/x);
        if(i<n) s+=v[i]+v[i];
        else
        s+=((pow(x,i/n))*v[i]);
       }
        else
        {
            if(i<n)
            {
                for(ll j=i;j<n;j++)
                s+=v[j];
                
            }
            
            break;
        }
    }
    cout<<s<<endl;
    
   
         
     }

int main()
{
          ios_base::sync_with_stdio(false);
          cin.tie(0);
          cout.tie();
          int t;
          cin>>t;
          //t=1;
          while(t--)
          {
                    solve();
          }
          return 0;
}
```