#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const ll MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;

struct node {
    int now;
    ll val;
    ll sum;
    bool friend operator < (node a, node b) {
        return a.now > b.now;
    }
};
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);


    int t;
    cin >> t;
    while(t--) {
        ll sum = 0;
        int n, x;
        cin >> n >> x;
        vector<int> ve;
        int minn = -1;
        for(int i = 1; i <= n; i++) {
            int k;
            cin >> k;
            int now = 0;
            ve.push_back(k);
            while(k % x == 0) {
                now++;
                k /= x;
            }
            if(minn == -1 || (now * n + i) < minn) {
                minn = (now * n + i);
            }
        }
        for(int i = 0; i < n; i++) {
            int now = 0;
            int tmp = ve[i];
            sum += tmp;
            while((now * n + i + 1) < minn) {
                sum += tmp;
                now++;
            }
        }
        cout << sum << endl;
    }

    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;
typedef unsigned long long ll;
const ll MOD = 1e9 + 7;
const int MAXN = 2e5 + 5;

struct node {
    int now;
    ll val;
    ll sum;
    bool friend operator < (node a, node b) {
        return a.now > b.now;
    }
};
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);


    int t;
    cin >> t;
    while(t--) {
        ll sum = 0;
        int n, x;
        cin >> n >> x;
        vector<int> ve;
        int minn = -1;
        for(int i = 1; i <= n; i++) {
            int k;
            cin >> k;
            int now = 0;
            ve.push_back(k);
            while(k % x == 0) {
                now++;
                k /= x;
            }
            if(minn == -1 || (now * n + i) < minn) {
                minn = (now * n + i);
            }
        }
        for(int i = 0; i < n; i++) {
            int now = 0;
            int tmp = ve[i];
            sum += tmp;
            while((now * n + i + 1) < minn) {
                sum += tmp;
                now++;
            }
        }
        cout << sum << endl;
    }

    return 0;
}
```