#include <bits/stdc++.h>

using namespace std;

void testCase()
{
	long long n, x;
	cin >> n >> x;
	deque<pair<long long, long long>> a(n);
	for (long long i = 0, ai; i < n; i++)
	{
		cin >> ai;
		a.push_back({ai, 1});
	}
	long long sum = 0;
	for (long long ai; !a.empty() && (ai = a.front().first) % x == 0; a.pop_front())
	{
		a.push_back({ai / x, a.front().second * x});
		sum += ai * a.front().second;
	}
	for (long long i = 0; i < a.size(); i++)
	{
		sum += a[i].first * a[i].second;
	}
	cout << sum << '\n';
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int t;
	cin >> t;
	while (t--)
		testCase();
}
### Prompt

Please provide a Cpp coded solution 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
#include <bits/stdc++.h>

using namespace std;

void testCase()
{
	long long n, x;
	cin >> n >> x;
	deque<pair<long long, long long>> a(n);
	for (long long i = 0, ai; i < n; i++)
	{
		cin >> ai;
		a.push_back({ai, 1});
	}
	long long sum = 0;
	for (long long ai; !a.empty() && (ai = a.front().first) % x == 0; a.pop_front())
	{
		a.push_back({ai / x, a.front().second * x});
		sum += ai * a.front().second;
	}
	for (long long i = 0; i < a.size(); i++)
	{
		sum += a[i].first * a[i].second;
	}
	cout << sum << '\n';
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int t;
	cin >> t;
	while (t--)
		testCase();
}
```