#include <bits/stdc++.h>
#include <string>
#include <vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<math.h>
#include <bits/stdc++.h>
using namespace std;
bool sortbysecdesc(const pair<int,int> &a,
                   const pair<int,int> &b)
{
       return a.second>b.second;
}
int main()
{
  	ios_base::sync_with_stdio(false);
    cin.tie(NULL); 
   		long long z;
		   cin>>z;
		while(z--)
		{
		long long i,j,k,l=0,r=0,n,mid=0,m,x,y,s1=0,s2=INT_MAX,s3=0,s5=0,s6=0,s4=0,p1=0,p2=0;
		cin>>n>>m;
		long long int a[n];
		for(i=0;i<n;i++)
		{
			cin>>a[i];
			x=a[i];
			if(a[i]%m==0)
			{
			s1=1;
			j=m;
			while(x%j==0)
			{
				s1++;
				x/=j;
			}
			if(s2>s1)
			{
				s2=min(s2,s1);
				s3=i;
			} 
			}
			else if(s2>1)
			{
				s2=1;
				s3=i;
			} 
		}
		s4=0;
		for(i=0;i<n;i++)
		{
			if(i<s3)
			s4+=((s2+1)*a[i]);
			else
			s4+=(s2*a[i]);
		}
		cout<<s4<<"\n";
	}		
}

### 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>
#include <string>
#include <vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<math.h>
#include <bits/stdc++.h>
using namespace std;
bool sortbysecdesc(const pair<int,int> &a,
                   const pair<int,int> &b)
{
       return a.second>b.second;
}
int main()
{
  	ios_base::sync_with_stdio(false);
    cin.tie(NULL); 
   		long long z;
		   cin>>z;
		while(z--)
		{
		long long i,j,k,l=0,r=0,n,mid=0,m,x,y,s1=0,s2=INT_MAX,s3=0,s5=0,s6=0,s4=0,p1=0,p2=0;
		cin>>n>>m;
		long long int a[n];
		for(i=0;i<n;i++)
		{
			cin>>a[i];
			x=a[i];
			if(a[i]%m==0)
			{
			s1=1;
			j=m;
			while(x%j==0)
			{
				s1++;
				x/=j;
			}
			if(s2>s1)
			{
				s2=min(s2,s1);
				s3=i;
			} 
			}
			else if(s2>1)
			{
				s2=1;
				s3=i;
			} 
		}
		s4=0;
		for(i=0;i<n;i++)
		{
			if(i<s3)
			s4+=((s2+1)*a[i]);
			else
			s4+=(s2*a[i]);
		}
		cout<<s4<<"\n";
	}		
}

```