#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
    ios_base::sync_with_stdio(0) ;
    cin.tie(0) ;
    cout.tie(0) ;
    int t;
    cin>>t;
    while(t--)
    {
        ll n,x,s=0,a=1;
        cin>>n>>x;
        ll arr[n];
        for(int i=0; i<n; i++)
        {
            cin>>arr[i];
        }

        int mn=100000000,gg=mn,ind=-1;

        for(int i=0; i<n; i++)
        {
            int y=arr[i],cnt=0;
            while(!(y%x))
            {
                y/=x;
                cnt++;
            }
            if(cnt<mn)
            {
                mn=cnt;
                ind=i;
            }
        }

        for(int i=0; i<n; i++)
        {
            if(i<ind)
            gg=mn+1;
            else
                gg=mn;
            s+=arr[i];
            a=1;
            while(((arr[i]%x)==0)&&(gg--))
            {
                a*=x;
                arr[i]/=x;
                s+=a*arr[i];
            }
        }

        cout<<s<<"\n";

    }

}

### Prompt

In CPP, your task is to solve 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>
#define ll long long
using namespace std;

int main()
{
    ios_base::sync_with_stdio(0) ;
    cin.tie(0) ;
    cout.tie(0) ;
    int t;
    cin>>t;
    while(t--)
    {
        ll n,x,s=0,a=1;
        cin>>n>>x;
        ll arr[n];
        for(int i=0; i<n; i++)
        {
            cin>>arr[i];
        }

        int mn=100000000,gg=mn,ind=-1;

        for(int i=0; i<n; i++)
        {
            int y=arr[i],cnt=0;
            while(!(y%x))
            {
                y/=x;
                cnt++;
            }
            if(cnt<mn)
            {
                mn=cnt;
                ind=i;
            }
        }

        for(int i=0; i<n; i++)
        {
            if(i<ind)
            gg=mn+1;
            else
                gg=mn;
            s+=arr[i];
            a=1;
            while(((arr[i]%x)==0)&&(gg--))
            {
                a*=x;
                arr[i]/=x;
                s+=a*arr[i];
            }
        }

        cout<<s<<"\n";

    }

}

```