/******************************************************************
Copyright: 11D_Beyonder All Rights Reserved
Author: 11D_Beyonder
Problem ID: CF1471B
Date: 01/13/2020
Description: Greedy
*******************************************************************/
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100006;
int _;
int n,x;
int dep[N],a[N];
int main(){
	for(cin>>_;_;_--){
		cin>>n>>x;
		ll ans=0;
		for(int i=1;i<=n;i++){
			cin>>a[i];
			dep[i]=1;
			ll q=a[i];
			while(q%x==0){
				dep[i]++;
				q/=x;
			}
		}
		int pos,MinDep=0x3f3f3f3f;
		for(int i=1;i<=n;i++){
			if(MinDep>dep[i]){
				MinDep=dep[i];
				pos=i;
			}
		}
		for(int i=1;i<pos;i++){
			ans+=1ll*(MinDep+1)*a[i];
		}
		for(int i=pos;i<=n;i++){
			ans+=1ll*MinDep*a[i];
		}
		cout<<ans<<endl;
	}
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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
/******************************************************************
Copyright: 11D_Beyonder All Rights Reserved
Author: 11D_Beyonder
Problem ID: CF1471B
Date: 01/13/2020
Description: Greedy
*******************************************************************/
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100006;
int _;
int n,x;
int dep[N],a[N];
int main(){
	for(cin>>_;_;_--){
		cin>>n>>x;
		ll ans=0;
		for(int i=1;i<=n;i++){
			cin>>a[i];
			dep[i]=1;
			ll q=a[i];
			while(q%x==0){
				dep[i]++;
				q/=x;
			}
		}
		int pos,MinDep=0x3f3f3f3f;
		for(int i=1;i<=n;i++){
			if(MinDep>dep[i]){
				MinDep=dep[i];
				pos=i;
			}
		}
		for(int i=1;i<pos;i++){
			ans+=1ll*(MinDep+1)*a[i];
		}
		for(int i=pos;i<=n;i++){
			ans+=1ll*MinDep*a[i];
		}
		cout<<ans<<endl;
	}
	return 0;
}
```