#include <bits/stdc++.h>
#define ls i << 1
#define rs i << 1 | 1
#define lowbit(x) (x & (-x))
#define inf 0x3f3f3f3f
#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mes(a, b) memset((a), (b), sizeof(a))
#define debug cout << "*****" << endl
#define pb push_back
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
const ll maxx = 1e5 + 10;
const ll mod = 1e9 + 7;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
ll s[maxx];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        ll sum=0;
        int n,x;
        cin>>n>>x;
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            sum+=s[i];
        }
        ll pos=x;
        for(int i=0;i<n;i++)
        {
            if(s[i]%pos==0)
                sum+=s[i];
            else
                break;
            if(i==n-1)
                pos*=x,i=-1;
        }
        cout<<sum<<endl;
    }
    return 0;
}

### 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 ls i << 1
#define rs i << 1 | 1
#define lowbit(x) (x & (-x))
#define inf 0x3f3f3f3f
#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define mes(a, b) memset((a), (b), sizeof(a))
#define debug cout << "*****" << endl
#define pb push_back
#define PI acos(-1.0)
using namespace std;
typedef long long ll;
const ll maxx = 1e5 + 10;
const ll mod = 1e9 + 7;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
ll s[maxx];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        ll sum=0;
        int n,x;
        cin>>n>>x;
        for(int i=0;i<n;i++)
        {
            cin>>s[i];
            sum+=s[i];
        }
        ll pos=x;
        for(int i=0;i<n;i++)
        {
            if(s[i]%pos==0)
                sum+=s[i];
            else
                break;
            if(i==n-1)
                pos*=x,i=-1;
        }
        cout<<sum<<endl;
    }
    return 0;
}

```