#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
map<ll,ll> m;
ll t,n,x,sum,k,minx,answer;
 
int main(){
	ll a[100005],d[100005];
	cin>>t;
	while(t--){
		sum=0;
		minx=99999;
		cin>>n>>x;
		for(int i=0;i<n;i++){
			cin>>a[i];
			sum+=a[i];
			d[i]=0;
		}
		
		for(int i=0;i<n;i++){
			k=a[i];
			if(m[k]!=0) d[i]=m[k];
			else
			if(k%x==0)	
			{
				while(k%x==0){
					d[i]++;
					k/=x;
				}
				m[a[i]]=d[i];
			}
			else break;
		}
		for(int i=0;i<n;i++)
		if(d[i]<minx) minx=d[i];
		answer=(minx+1)*sum;
		for(int i=0;i<n;i++){
			d[i]-=minx;
			if(d[i]==0) break;
			else answer+=a[i];
		}
		cout<<answer<<endl;
		m.clear();
	}
}
### 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;
 
typedef long long ll;
map<ll,ll> m;
ll t,n,x,sum,k,minx,answer;
 
int main(){
	ll a[100005],d[100005];
	cin>>t;
	while(t--){
		sum=0;
		minx=99999;
		cin>>n>>x;
		for(int i=0;i<n;i++){
			cin>>a[i];
			sum+=a[i];
			d[i]=0;
		}
		
		for(int i=0;i<n;i++){
			k=a[i];
			if(m[k]!=0) d[i]=m[k];
			else
			if(k%x==0)	
			{
				while(k%x==0){
					d[i]++;
					k/=x;
				}
				m[a[i]]=d[i];
			}
			else break;
		}
		for(int i=0;i<n;i++)
		if(d[i]<minx) minx=d[i];
		answer=(minx+1)*sum;
		for(int i=0;i<n;i++){
			d[i]-=minx;
			if(d[i]==0) break;
			else answer+=a[i];
		}
		cout<<answer<<endl;
		m.clear();
	}
}
```