#include <bits/stdc++.h>
using namespace std;
const int INF = 1e7 + 7;
int k[100005];
int main()
{
  int n, a, x, p, l;
  cin >> n;
  while (n--)
  {
    long long t = 0;
    cin >> a >> x;
    int Min = INF, d;
    for (int i = 0; i < a; i++)
    {
      scanf("%d", &k[i]);
      l = 0, d = k[i];
      while (!(d % x))
      {
        l++;
        d /= x;
      }
      if (l < Min)
      {
        Min = l;
        p = i;
      }
      t += k[i];
    }
    t *= (Min + 1);
    for (int i = 0; i < p; i++)
      t += k[i];
    cout << t << endl;
  }
  return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;
const int INF = 1e7 + 7;
int k[100005];
int main()
{
  int n, a, x, p, l;
  cin >> n;
  while (n--)
  {
    long long t = 0;
    cin >> a >> x;
    int Min = INF, d;
    for (int i = 0; i < a; i++)
    {
      scanf("%d", &k[i]);
      l = 0, d = k[i];
      while (!(d % x))
      {
        l++;
        d /= x;
      }
      if (l < Min)
      {
        Min = l;
        p = i;
      }
      t += k[i];
    }
    t *= (Min + 1);
    for (int i = 0; i < p; i++)
      t += k[i];
    cout << t << endl;
  }
  return 0;
}
```