#include <bits/stdc++.h>
#define FLASH ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define all(x) x.begin(),x.end() 
#define int long long int
#define double long double
#define MAX 1e9
#define MIN -1e9
#define mod 1000000007
#define mxn 100005
#define vi vector<int>
#define pb push_back
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define output(x) cout << ( x ? "YES" : "NO" ) << '\n' ;
using namespace std;
void solve()
{
    int n,x;
    cin>>n>>x;
    vector <pair<int,int>> a;
    int ter=1,sum=0,s=0;
    for(int i=0;i<n;i++)
    {
        int m;
        cin>>m;
        sum+=m;
        if(m%x==0 && ter)
        {
            a.pb(make_pair(m/x,x));
            s++;
        }
        else
            ter=0;
    }
    for(int i=0;i<s;i++)
    {
        int m=a[i].first;
        sum+=m*(a[i].second);
        if(m%x==0 && ter)
        {
            a.pb(make_pair(m/x,a[i].second*x));
            s++;
        }
        else
        {
            ter=0;
        }
    }
    cout<<sum<<'\n';
}
signed main()
{
    FLASH
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}

//cout<<"###";
//cout<<"***";
### 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>
#define FLASH ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define all(x) x.begin(),x.end() 
#define int long long int
#define double long double
#define MAX 1e9
#define MIN -1e9
#define mod 1000000007
#define mxn 100005
#define vi vector<int>
#define pb push_back
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define output(x) cout << ( x ? "YES" : "NO" ) << '\n' ;
using namespace std;
void solve()
{
    int n,x;
    cin>>n>>x;
    vector <pair<int,int>> a;
    int ter=1,sum=0,s=0;
    for(int i=0;i<n;i++)
    {
        int m;
        cin>>m;
        sum+=m;
        if(m%x==0 && ter)
        {
            a.pb(make_pair(m/x,x));
            s++;
        }
        else
            ter=0;
    }
    for(int i=0;i<s;i++)
    {
        int m=a[i].first;
        sum+=m*(a[i].second);
        if(m%x==0 && ter)
        {
            a.pb(make_pair(m/x,a[i].second*x));
            s++;
        }
        else
        {
            ter=0;
        }
    }
    cout<<sum<<'\n';
}
signed main()
{
    FLASH
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}

//cout<<"###";
//cout<<"***";
```