#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<utility>
#include<vector>
#include<stack>
#include<queue>
#include<cstring>
#include<set>
#include<map>
#define endl "\n"
#define all(v) v.begin(),v.end()
#define st first
#define nd second
#define mp make_pair
#define pb push_back
using namespace std;
typedef long long lo;
const int mod=1000000007,N=500005;
lo a,b,c,d,e,f,g=1,h[N],arr[N];
string s;
vector<pair<lo,lo> >v;
 
void solve(){
	v.clear();
	cin >> a >> b;
	lo sum=0;
	bool ok=0;
	for(lo i=1;i<=a;i++){
		cin >> arr[i];
		sum+=arr[i];
		if(arr[i]%b!=0){
			ok=1;
		}
		if(ok==0){
			v.pb({arr[i]/b,b});
			sum+=arr[i];
		}
	}
	if(ok){cout << sum << endl; return;}
	for(lo i=0;i<v.size();i++){
		if(v[i].st%b!=0){
			cout << sum << endl;
			return;
		}
		sum+=v[i].st*v[i].nd;
		v.pb({v[i].st/b,b*v[i].nd});
	}
}
 
int main(){
	#ifdef local
		freopen("in.txt","r",stdin);
		freopen("out.txt","w",stdout);
	#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	cin >> g;
	while(g--)solve();
}
### Prompt

Generate 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<cstdio>
#include<algorithm>
#include<cmath>
#include<utility>
#include<vector>
#include<stack>
#include<queue>
#include<cstring>
#include<set>
#include<map>
#define endl "\n"
#define all(v) v.begin(),v.end()
#define st first
#define nd second
#define mp make_pair
#define pb push_back
using namespace std;
typedef long long lo;
const int mod=1000000007,N=500005;
lo a,b,c,d,e,f,g=1,h[N],arr[N];
string s;
vector<pair<lo,lo> >v;
 
void solve(){
	v.clear();
	cin >> a >> b;
	lo sum=0;
	bool ok=0;
	for(lo i=1;i<=a;i++){
		cin >> arr[i];
		sum+=arr[i];
		if(arr[i]%b!=0){
			ok=1;
		}
		if(ok==0){
			v.pb({arr[i]/b,b});
			sum+=arr[i];
		}
	}
	if(ok){cout << sum << endl; return;}
	for(lo i=0;i<v.size();i++){
		if(v[i].st%b!=0){
			cout << sum << endl;
			return;
		}
		sum+=v[i].st*v[i].nd;
		v.pb({v[i].st/b,b*v[i].nd});
	}
}
 
int main(){
	#ifdef local
		freopen("in.txt","r",stdin);
		freopen("out.txt","w",stdout);
	#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	cin >> g;
	while(g--)solve();
}
```