#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;


int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout.setf(ios::fixed); cout.precision(20);
    // freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        ll x; cin >> x;
        queue<pll> q;
        for (int i = 0; i < n; ++i) {
            ll r; cin >> r;
            q.emplace(1, r);
        }

        ll sm = 0;
        while (!q.empty()) {
            auto [am, num] = q.front();
            q.pop();

            sm += num * am;

            if (num % x != 0) {
                break;
            } else {
                q.emplace(am * x, num / x);
            }
        }

        while (!q.empty()) {
            auto [am, num] = q.front();
            q.pop();
            sm += num * am;
        }

        cout << sm << endl;

    }

    return 0;
}
### 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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;


int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout.setf(ios::fixed); cout.precision(20);
    // freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        ll x; cin >> x;
        queue<pll> q;
        for (int i = 0; i < n; ++i) {
            ll r; cin >> r;
            q.emplace(1, r);
        }

        ll sm = 0;
        while (!q.empty()) {
            auto [am, num] = q.front();
            q.pop();

            sm += num * am;

            if (num % x != 0) {
                break;
            } else {
                q.emplace(am * x, num / x);
            }
        }

        while (!q.empty()) {
            auto [am, num] = q.front();
            q.pop();
            sm += num * am;
        }

        cout << sm << endl;

    }

    return 0;
}
```