#include <iostream>
#include <algorithm>
#define ll long long 
using namespace std;

ll tcases;
ll factors[200010];
ll basearr[200010];

int main()
{
	cin >> tcases;
	for (int t = 0; t < tcases; t++) {
		int length;
		int x;
		cin >> length >> x;
		for (int i = 0; i < length; i++) {
			cin >> basearr[i];
			factors[i] = 0;
		}
		for (int i = 0; i < length; i++) {
			int t = basearr[i];
			while (t%x == 0) {
				t /= x;
				factors[i]++;
			}
		}
		int ming = 10000;
		int mindex = 0;
		for (int i = 0; i < length; i++) {
			if (factors[i] < ming) {
				ming = factors[i];
				mindex = i;
			}
		}
		long long ans = 0;
		for (int i = 0; i < mindex; i++) {
			ans += basearr[i] * (ming+2);
		}
		for (int i = mindex; i < length; i++) {
			ans += basearr[i] * (ming+1);
		}
		cout << ans << endl;
	}
}
### 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 <iostream>
#include <algorithm>
#define ll long long 
using namespace std;

ll tcases;
ll factors[200010];
ll basearr[200010];

int main()
{
	cin >> tcases;
	for (int t = 0; t < tcases; t++) {
		int length;
		int x;
		cin >> length >> x;
		for (int i = 0; i < length; i++) {
			cin >> basearr[i];
			factors[i] = 0;
		}
		for (int i = 0; i < length; i++) {
			int t = basearr[i];
			while (t%x == 0) {
				t /= x;
				factors[i]++;
			}
		}
		int ming = 10000;
		int mindex = 0;
		for (int i = 0; i < length; i++) {
			if (factors[i] < ming) {
				ming = factors[i];
				mindex = i;
			}
		}
		long long ans = 0;
		for (int i = 0; i < mindex; i++) {
			ans += basearr[i] * (ming+2);
		}
		for (int i = mindex; i < length; i++) {
			ans += basearr[i] * (ming+1);
		}
		cout << ans << endl;
	}
}
```