#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define int ll
const int mxN = (int)1e5+5;
int n, x, ar[mxN];

void solve() {
	cin >> n >> x;
	int sum = 0;
	int ar2[n];
	for (int i = 0; i < n; i++) {
		cin >> ar[i];
		sum += ar[i];
		ar2[i] = ar[i];
	}

	int i = 0;
	while(true) {
		if (ar[i]%x != 0)
			break;
		
		sum += ar2[i];
		ar[i]/=x;

		i++;
		if (i == n)
			i = 0;
	}
	cout << sum << "\n";
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t = 1, i = 1;
	cin >> t;
	while(t--) {
		//cout << "Case #" << i << ": ";
		solve();
		//i++;
	}
	return 0;
}


### Prompt

Please create a solution in CPP 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;

#define ll long long
#define int ll
const int mxN = (int)1e5+5;
int n, x, ar[mxN];

void solve() {
	cin >> n >> x;
	int sum = 0;
	int ar2[n];
	for (int i = 0; i < n; i++) {
		cin >> ar[i];
		sum += ar[i];
		ar2[i] = ar[i];
	}

	int i = 0;
	while(true) {
		if (ar[i]%x != 0)
			break;
		
		sum += ar2[i];
		ar[i]/=x;

		i++;
		if (i == n)
			i = 0;
	}
	cout << sum << "\n";
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t = 1, i = 1;
	cin >> t;
	while(t--) {
		//cout << "Case #" << i << ": ";
		solve();
		//i++;
	}
	return 0;
}


```