#include <iostream>
#include <cmath>
#include <string>
#include <array>
#include <set>
#include <algorithm>
#include <map>
#include <cstring>
#include <vector>
#include <iomanip>
#include <array>
#include <stack>
#include <queue>
#include <deque>
#include <numeric>
typedef long long ll;

using namespace std;
ll a[1000000],b[1000001],c[100001], n,m,k,ans = 0,ans1 = 0,x ,y ,z,t ,h,v = 0,mx=0,mn = 0,sum = 0,cnt = 0;

int main() {
    cin>>t;
    while(t--){
        cin>>n>>m;
        sum = 0;

        for (int i = 0; i < n; ++i) {
            cin>>a[i];
            b[i] = a[i];
            sum += a[i];
        }

        bool ck = true;
        while(ck){
            for (int i = 0; i < n; ++i) {
                if(b[i] % m == 0) sum+=a[i],b[i] /=m;
                else{
                    ck = false;
                    break;
                }
            }
        }

        cout<<sum<<endl;

    }
}
### Prompt

Your task is to create 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 <iostream>
#include <cmath>
#include <string>
#include <array>
#include <set>
#include <algorithm>
#include <map>
#include <cstring>
#include <vector>
#include <iomanip>
#include <array>
#include <stack>
#include <queue>
#include <deque>
#include <numeric>
typedef long long ll;

using namespace std;
ll a[1000000],b[1000001],c[100001], n,m,k,ans = 0,ans1 = 0,x ,y ,z,t ,h,v = 0,mx=0,mn = 0,sum = 0,cnt = 0;

int main() {
    cin>>t;
    while(t--){
        cin>>n>>m;
        sum = 0;

        for (int i = 0; i < n; ++i) {
            cin>>a[i];
            b[i] = a[i];
            sum += a[i];
        }

        bool ck = true;
        while(ck){
            for (int i = 0; i < n; ++i) {
                if(b[i] % m == 0) sum+=a[i],b[i] /=m;
                else{
                    ck = false;
                    break;
                }
            }
        }

        cout<<sum<<endl;

    }
}
```