#include <iostream>
#include <cmath>
#include <vector>
using namespace std;




int main() {
	// your code goes here
       
   int t;
    cin>>t;
    while(t>0)
    {
        int n,x;
        cin>>n>>x;
        int temp;
        vector<int> v;
        vector<int> u;
        long long int sum=0;
        bool y=0;
        for(int j=0;j<n;j++){cin>>temp;v.push_back(temp);u.push_back(temp);}
        for(int i =0;i<(int)v.size();i++)
        {
            sum+= v[i];
            if(v[i]%x==0&&u[i]%x==0&&y==0)
            {
                int j=x;
                int k= u[i] /x;
                v.push_back(v[i]);
                u.push_back(k);
            }
            else
            {
                y=1;
            }
            //cout<<"S "<<sum<<endl;
            //cout<<"V: ";for(int it =0;it<(int)v.size();it++){cout<<v[it]<<" ";}cout<<endl;
            //cout<<"U: ";for(int it =0;it<(int)u.size();it++){cout<<u[it]<<" ";}cout<<endl;
        }

        cout<<sum<<endl;
        t--;
    }
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <vector>
using namespace std;




int main() {
	// your code goes here
       
   int t;
    cin>>t;
    while(t>0)
    {
        int n,x;
        cin>>n>>x;
        int temp;
        vector<int> v;
        vector<int> u;
        long long int sum=0;
        bool y=0;
        for(int j=0;j<n;j++){cin>>temp;v.push_back(temp);u.push_back(temp);}
        for(int i =0;i<(int)v.size();i++)
        {
            sum+= v[i];
            if(v[i]%x==0&&u[i]%x==0&&y==0)
            {
                int j=x;
                int k= u[i] /x;
                v.push_back(v[i]);
                u.push_back(k);
            }
            else
            {
                y=1;
            }
            //cout<<"S "<<sum<<endl;
            //cout<<"V: ";for(int it =0;it<(int)v.size();it++){cout<<v[it]<<" ";}cout<<endl;
            //cout<<"U: ";for(int it =0;it<(int)u.size();it++){cout<<u[it]<<" ";}cout<<endl;
        }

        cout<<sum<<endl;
        t--;
    }
    return 0;
}

```