#include<bits/stdc++.h>

using namespace std;

#define endl "\n"
#define ll long long int

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    ll t;
    cin>>t;
    
    while(t--)
    {
        ll n,x;
        cin>>n>>x;
        
        vector<pair<ll,ll>> v;
        
        for(int i=0;i<n;i++)
        {
            ll k;
            cin>>k;
            v.push_back({k,1});
        }
        
        ll l=x;
        
        for(int i=1;i<=v.size();i++)
        {
            if(v[i-1].first%x!=0)
                break;
            v.push_back({(v[i-1].first/x),l});
            if(i%n==0)
               l*=x;
        }
        
        /*for(auto c:v)
            cout<<c.first<<" "<<c.second<<endl;*/
        
        ll sum=0;
        
        for(int i=0;i<v.size();i++)
        {
            sum+=(v[i].first*v[i].second);
        }
        
        cout<<sum<<endl;
        
    }
    
    return 0;
}











### 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<bits/stdc++.h>

using namespace std;

#define endl "\n"
#define ll long long int

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    ll t;
    cin>>t;
    
    while(t--)
    {
        ll n,x;
        cin>>n>>x;
        
        vector<pair<ll,ll>> v;
        
        for(int i=0;i<n;i++)
        {
            ll k;
            cin>>k;
            v.push_back({k,1});
        }
        
        ll l=x;
        
        for(int i=1;i<=v.size();i++)
        {
            if(v[i-1].first%x!=0)
                break;
            v.push_back({(v[i-1].first/x),l});
            if(i%n==0)
               l*=x;
        }
        
        /*for(auto c:v)
            cout<<c.first<<" "<<c.second<<endl;*/
        
        ll sum=0;
        
        for(int i=0;i<v.size();i++)
        {
            sum+=(v[i].first*v[i].second);
        }
        
        cout<<sum<<endl;
        
    }
    
    return 0;
}











```