#include<bits/stdc++.h>
#define lli long long int
using namespace std;
int main()
{ int t;
  cin>>t;
  while(t--)
  { lli n,x;
    cin>>n>>x;
    vector<lli> a(n),temp(n);
    lli Sum=0;
    for(int i=0;i<n;i++)
    { 
      cin>>a[i];
      Sum+=a[i];
    }
    int i=0,bit=1;
    while(!a[i]==0)
    { if(a[i]%x==0)
      { a[i]=a[i]/x;
        Sum+=pow(x,bit)*a[i];
        i++;
        if(i==n)
        {
          i=0;
          bit++;
        }
      }
      else
      {
         break; 
      }
    }
   cout<<Sum<<'\n'; 
  }
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 lli long long int
using namespace std;
int main()
{ int t;
  cin>>t;
  while(t--)
  { lli n,x;
    cin>>n>>x;
    vector<lli> a(n),temp(n);
    lli Sum=0;
    for(int i=0;i<n;i++)
    { 
      cin>>a[i];
      Sum+=a[i];
    }
    int i=0,bit=1;
    while(!a[i]==0)
    { if(a[i]%x==0)
      { a[i]=a[i]/x;
        Sum+=pow(x,bit)*a[i];
        i++;
        if(i==n)
        {
          i=0;
          bit++;
        }
      }
      else
      {
         break; 
      }
    }
   cout<<Sum<<'\n'; 
  }
}
```