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

int main()
{
    const long long M = 1e9+7 ;
    #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin) ;
     freopen("output.txt", "w", stdout) ;
    #endif

    ios_base::sync_with_stdio(false) ;
    cin.tie(NULL) ;
    cout.tie(NULL) ;

    int test=1; cin>>test;

    while(test--)
    {
        ll len, x;
        cin>>len>>x ;

        vector <ll> arr(len) ;
        ll sum = 0;
        int level=INT_MAX, ind=-1 ;
        for(int i=0; i<len; i++)
            {
                cin>>arr[i] ;
                sum += arr[i] ;
            }

        for(int i=0; i<len; i++)
        {
            int t = 0 ;
            ll temp = arr[i] ;

            while(temp % x == 0)
            {
                t++ ;
                temp = temp / x ;
            }

            if(level > t)
            {
                level = t ;
                ind = i ;
            }
        }

        for(int i=0; i<len; i++)
        {
            if(i<ind)
                sum += arr[i] * (level+1) ;

            if(i>=ind)
                sum += arr[i] * (level) ;
        }

        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 ll long long

int main()
{
    const long long M = 1e9+7 ;
    #ifndef ONLINE_JUDGE
     freopen("input.txt", "r", stdin) ;
     freopen("output.txt", "w", stdout) ;
    #endif

    ios_base::sync_with_stdio(false) ;
    cin.tie(NULL) ;
    cout.tie(NULL) ;

    int test=1; cin>>test;

    while(test--)
    {
        ll len, x;
        cin>>len>>x ;

        vector <ll> arr(len) ;
        ll sum = 0;
        int level=INT_MAX, ind=-1 ;
        for(int i=0; i<len; i++)
            {
                cin>>arr[i] ;
                sum += arr[i] ;
            }

        for(int i=0; i<len; i++)
        {
            int t = 0 ;
            ll temp = arr[i] ;

            while(temp % x == 0)
            {
                t++ ;
                temp = temp / x ;
            }

            if(level > t)
            {
                level = t ;
                ind = i ;
            }
        }

        for(int i=0; i<len; i++)
        {
            if(i<ind)
                sum += arr[i] * (level+1) ;

            if(i>=ind)
                sum += arr[i] * (level) ;
        }

        cout<<sum<<endl ;
    }

    return 0;
}

```