#include<bits/stdc++.h>

using namespace std;
#define ll long long
#define int ll
#define debug(v) cout << ">>" << #v << ':' << v << endl;
#define endl "\n"
#define pb push_back
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define umap unordered_map
#define pii pair<int , int>
#define all(v) v.begin() , v.end()   
#define ld long double
#define min3(a,b,c) min(a,min(b,c))
#define min4(a,b,c,d ) min(a , min(b , min(c,d)))
#define max3(a,b,c) max(a , max(b,c))
#define max4(a,b,c,d) max(max(a,b) , max(c,d))
                                                 
//sort(v.begin(), v.end(), 
//[](auto &left, auto &right) 
//{return left.second < right.second;});
void reader()
{
   #ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input.txt", "r", stdin);
    freopen("output.txt" , "w" , stdout);
   #endif
}

const int MOD = 1e9 + 7;
const int N = 3e5+5;
const int inf = 1e9;


void solve()
{
   int n , x;
   cin>>n>>x;
   vector<int> v(n);
   int sum=0;
   for(int i=0 ; i<n ; i++)cin>>v[i];
   for(int i=0 ; i<v.size() ; i++)
   {
       if(v[i]%x==0)v.pb( v[i]/x );
       else break;
   }
   for(int i=0 ; i<v.size() ; i++)
   {
       int ex = i/n;
       int val = pow(x,ex);
       val = val*v[i];
       sum+=val;
   }
   cout<<sum<<endl;
}


signed main() 
{
    IOS
    reader();
    int t;
    cin>>t;    
    for(int z=1 ; z<=t ; z++)
    {
        //cout<<"Case #"<<z<<": ";
        solve();
    }
    
}


### Prompt

Please provide a CPP coded solution 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>

using namespace std;
#define ll long long
#define int ll
#define debug(v) cout << ">>" << #v << ':' << v << endl;
#define endl "\n"
#define pb push_back
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define umap unordered_map
#define pii pair<int , int>
#define all(v) v.begin() , v.end()   
#define ld long double
#define min3(a,b,c) min(a,min(b,c))
#define min4(a,b,c,d ) min(a , min(b , min(c,d)))
#define max3(a,b,c) max(a , max(b,c))
#define max4(a,b,c,d) max(max(a,b) , max(c,d))
                                                 
//sort(v.begin(), v.end(), 
//[](auto &left, auto &right) 
//{return left.second < right.second;});
void reader()
{
   #ifndef ONLINE_JUDGE
    // for getting input from input.txt
    freopen("input.txt", "r", stdin);
    freopen("output.txt" , "w" , stdout);
   #endif
}

const int MOD = 1e9 + 7;
const int N = 3e5+5;
const int inf = 1e9;


void solve()
{
   int n , x;
   cin>>n>>x;
   vector<int> v(n);
   int sum=0;
   for(int i=0 ; i<n ; i++)cin>>v[i];
   for(int i=0 ; i<v.size() ; i++)
   {
       if(v[i]%x==0)v.pb( v[i]/x );
       else break;
   }
   for(int i=0 ; i<v.size() ; i++)
   {
       int ex = i/n;
       int val = pow(x,ex);
       val = val*v[i];
       sum+=val;
   }
   cout<<sum<<endl;
}


signed main() 
{
    IOS
    reader();
    int t;
    cin>>t;    
    for(int z=1 ; z<=t ; z++)
    {
        //cout<<"Case #"<<z<<": ";
        solve();
    }
    
}


```