#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(c) (c).begin(), (c).end()

int t, n, m, x;

void solve() {
    cin >> n >> x;
    bool one = false;
    ll res = 0;
    vector<pair<ll, ll>> a (n);
    for (auto& b : a) {
        cin >> b.first;
        b.second = 1;
    }
    bool c = false;
    for (int i = 0; i < a.size(); ++i) {
        if (!c && a[i].first % x == 0) {
            res += a[i].first * a[i].second;
            a.push_back({a[i].first/x, a[i].second*x});
        } else {
            c = true;
            res += a[i].first * a[i].second;
        }
    }
    cout << res << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie();
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

### 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;
#define ll long long
#define all(c) (c).begin(), (c).end()

int t, n, m, x;

void solve() {
    cin >> n >> x;
    bool one = false;
    ll res = 0;
    vector<pair<ll, ll>> a (n);
    for (auto& b : a) {
        cin >> b.first;
        b.second = 1;
    }
    bool c = false;
    for (int i = 0; i < a.size(); ++i) {
        if (!c && a[i].first % x == 0) {
            res += a[i].first * a[i].second;
            a.push_back({a[i].first/x, a[i].second*x});
        } else {
            c = true;
            res += a[i].first * a[i].second;
        }
    }
    cout << res << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie();
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

```