#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll n,x;
        cin>>n>>x;
        // vector<int>v;
        
        queue<pair<ll,ll>>q;
        
        for(int i=0;i<n;++i)
        {
            ll ask;
            cin>>ask;
            q.push(make_pair(ask,1));
        }
        
        int flag=0;
        ll sum1=0;
        // int prev = 0;
        while(!q.empty())
        {
            pair<ll,ll>num1 = q.front();
            q.pop();
            
            // cout<<num1.first<<" "<<num1.second<<endl;
            
            
            if(num1.first%x==0 && flag==0)
            {
                q.push(make_pair(num1.first/x,x*num1.second));
            }
            else{
                flag=1;
            }
            
            
            
            sum1+=(num1.first*num1.second);
            
        }
        
        cout<<sum1<<endl;
        
    }
    return 0;
}
### Prompt

Create a solution in Cpp for 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 int ll;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll n,x;
        cin>>n>>x;
        // vector<int>v;
        
        queue<pair<ll,ll>>q;
        
        for(int i=0;i<n;++i)
        {
            ll ask;
            cin>>ask;
            q.push(make_pair(ask,1));
        }
        
        int flag=0;
        ll sum1=0;
        // int prev = 0;
        while(!q.empty())
        {
            pair<ll,ll>num1 = q.front();
            q.pop();
            
            // cout<<num1.first<<" "<<num1.second<<endl;
            
            
            if(num1.first%x==0 && flag==0)
            {
                q.push(make_pair(num1.first/x,x*num1.second));
            }
            else{
                flag=1;
            }
            
            
            
            sum1+=(num1.first*num1.second);
            
        }
        
        cout<<sum1<<endl;
        
    }
    return 0;
}
```