#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;

int T, n;
ll x, cnt[maxn], a[maxn];

int main() {
#ifndef ONLINE_JUDGE
	freopen("1.txt", "r", stdin);
#endif
	scanf("%d", &T);
	while (T--) {
		scanf("%d%lld", &n, &x);
		ll mn = 1e9, res = 0;
		for (int i = 1; i <= n; ++i) {
			scanf("%lld", &a[i]);
			ll tmp = a[i];
			cnt[i] = 0;
			while (tmp % x == 0) {
				++cnt[i];
				tmp /= x;
			}
			mn = min(mn, cnt[i] + 1);
		}
		bool flag = false;
		for (int i = 1; i <= n; ++i) {
			if (flag) {
				res += mn * a[i];
				continue;
			}
			if (cnt[i] + 1 > mn) {
				res += (mn + 1) * a[i];
			} else {
				res += mn * a[i];
				flag = true;
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}
### Prompt

Create a solution in Cpp for 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;
typedef long long ll;
const int maxn = 1e5 + 10;

int T, n;
ll x, cnt[maxn], a[maxn];

int main() {
#ifndef ONLINE_JUDGE
	freopen("1.txt", "r", stdin);
#endif
	scanf("%d", &T);
	while (T--) {
		scanf("%d%lld", &n, &x);
		ll mn = 1e9, res = 0;
		for (int i = 1; i <= n; ++i) {
			scanf("%lld", &a[i]);
			ll tmp = a[i];
			cnt[i] = 0;
			while (tmp % x == 0) {
				++cnt[i];
				tmp /= x;
			}
			mn = min(mn, cnt[i] + 1);
		}
		bool flag = false;
		for (int i = 1; i <= n; ++i) {
			if (flag) {
				res += mn * a[i];
				continue;
			}
			if (cnt[i] + 1 > mn) {
				res += (mn + 1) * a[i];
			} else {
				res += mn * a[i];
				flag = true;
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}
```