#include<iostream>
#include<bits/stdc++.h>
using namespace std;

#define ull unsigned long long int
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define pb push_back
void solve()
{
	int n,x;
    ull sum=0;
    cin>>n>>x;
    vector<int> a(n,0);
    for(int i=0;i<n;++i){
        cin>>a[i];
        sum+=a[i];
    }
    int i=0;
    vector<int> b(a);
    while(1){
        i=i%n;
        if(a[i]%x!=0) break;
        sum+=b[i];
        a[i]/=x;
        ++i;
    }
    
    cout<<sum<<endl;

}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
		solve();
	return  0;

}

/* another solution with less space complexity
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
 
#define ull unsigned long long int
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define pb push_back
void solve()
{
	int n,y,idx=0;
    ll sum=0,mini=1e9;
    cin>>n>>y;
    vector<int> a(n,0);
    for(int i=0;i<n;++i){
        ll x,cnt=0;
        cin>>x;
        a[i]=x;
        sum+=x;
        while(x%y==0){
            x=x/y;
            ++cnt;
        }
        
        if(cnt<mini){
            mini=cnt;
            idx=i;
        }
 
    }
    sum=sum*(mini+1LL);
    for(int i=0;i<idx;++i){
        sum+=a[i];
    }    
    cout<<sum<<endl;
 
}
 
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
		solve();
	return  0;
 
}
*/
### Prompt

Your challenge is to write 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<iostream>
#include<bits/stdc++.h>
using namespace std;

#define ull unsigned long long int
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define pb push_back
void solve()
{
	int n,x;
    ull sum=0;
    cin>>n>>x;
    vector<int> a(n,0);
    for(int i=0;i<n;++i){
        cin>>a[i];
        sum+=a[i];
    }
    int i=0;
    vector<int> b(a);
    while(1){
        i=i%n;
        if(a[i]%x!=0) break;
        sum+=b[i];
        a[i]/=x;
        ++i;
    }
    
    cout<<sum<<endl;

}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
		solve();
	return  0;

}

/* another solution with less space complexity
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
 
#define ull unsigned long long int
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define pb push_back
void solve()
{
	int n,y,idx=0;
    ll sum=0,mini=1e9;
    cin>>n>>y;
    vector<int> a(n,0);
    for(int i=0;i<n;++i){
        ll x,cnt=0;
        cin>>x;
        a[i]=x;
        sum+=x;
        while(x%y==0){
            x=x/y;
            ++cnt;
        }
        
        if(cnt<mini){
            mini=cnt;
            idx=i;
        }
 
    }
    sum=sum*(mini+1LL);
    for(int i=0;i<idx;++i){
        sum+=a[i];
    }    
    cout<<sum<<endl;
 
}
 
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--)
		solve();
	return  0;
 
}
*/
```