#include <bits/stdc++.h>
using namespace std;
const int MM = 1e5 + 5;
long long t, n, A[MM], x, cnt = INT_MAX, idx0 = -1;

int main() {
    cin.tie(0);
    cin >> t;
    while(t--) {
        cnt = INT_MAX, idx0 = -1;
        cin >> n >> x;
        for (int i = 1; i <= n; i++) cin >> A[i];
        for (int i = 1; i <= n; i++) {
            int tmp = A[i], tcnt = 0;
            while (tmp % x == 0) tmp /= x, tcnt++;
            if (tcnt < cnt) {
                cnt = tcnt;
                idx0 = i;
            }
        }
        long long sum = 0;
        for (int i = 1; i < idx0; i++) sum += (cnt + 2) * A[i];
        for (int i = idx0; i <= n; i++) sum += (cnt + 1) * A[i];
        cout << sum << "\n";
    }
}
### Prompt

Develop a solution in Cpp 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;
const int MM = 1e5 + 5;
long long t, n, A[MM], x, cnt = INT_MAX, idx0 = -1;

int main() {
    cin.tie(0);
    cin >> t;
    while(t--) {
        cnt = INT_MAX, idx0 = -1;
        cin >> n >> x;
        for (int i = 1; i <= n; i++) cin >> A[i];
        for (int i = 1; i <= n; i++) {
            int tmp = A[i], tcnt = 0;
            while (tmp % x == 0) tmp /= x, tcnt++;
            if (tcnt < cnt) {
                cnt = tcnt;
                idx0 = i;
            }
        }
        long long sum = 0;
        for (int i = 1; i < idx0; i++) sum += (cnt + 2) * A[i];
        for (int i = idx0; i <= n; i++) sum += (cnt + 1) * A[i];
        cout << sum << "\n";
    }
}
```