#include <bits/stdc++.h>
using namespace std;
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n"

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		//task create acylic

		long long int n,x;
		cin>>n>>x;
		long long int sm=0;
		vector<long long int> v(n),cp(n);
		for(int i=0;i<n;i++)
		{
			cin>>v[i];
			cp[i]=v[i];
		}
		while(1)
		{	
			int flag=0,sp=INT_MAX;
			for(int i=0;i<n;i++)
			{
				if(cp[i]%x==0)
				{
					sm+=v[i];
					cp[i]=cp[i]/x;
					
					
				}
				else
				{
					sm+=v[i];
					flag=1;
					sp=min(sp,i);
					// break;
				}
			}
			// cout<<sm<<endl;
			if(flag)
			{
				for(int i=0;i<sp;i++)
					sm+=v[i];
				break;
			}
		}
		cout<<sm<<endl;
	}

}
### Prompt

Create a solution in Cpp for 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 time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n"

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		//task create acylic

		long long int n,x;
		cin>>n>>x;
		long long int sm=0;
		vector<long long int> v(n),cp(n);
		for(int i=0;i<n;i++)
		{
			cin>>v[i];
			cp[i]=v[i];
		}
		while(1)
		{	
			int flag=0,sp=INT_MAX;
			for(int i=0;i<n;i++)
			{
				if(cp[i]%x==0)
				{
					sm+=v[i];
					cp[i]=cp[i]/x;
					
					
				}
				else
				{
					sm+=v[i];
					flag=1;
					sp=min(sp,i);
					// break;
				}
			}
			// cout<<sm<<endl;
			if(flag)
			{
				for(int i=0;i<sp;i++)
					sm+=v[i];
				break;
			}
		}
		cout<<sm<<endl;
	}

}
```