// Problem: B. Strange List
// Contest: Codeforces - Codeforces Round #694 (Div. 2)
// URL: https://codeforces.com/contest/1471/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms

//code by: Ravi Khatri @Ravi_Orz
#include<bits/stdc++.h>
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define F first
#define S second
#define pb push_back
#define f(i,s,n) for(int i=(int)s;i<=(int)n;++i)
#define fr(i,n,s) for(int i=(int)n;i>=s;--i)
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define Fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;
#define mod 1000000007
ll n,m,l,r,x;
ll a[1000006];
ll b[1000006];
vi g[1000006];

void solve()
{	ll sum=0;
	cin>>n>>x;
	queue< pair<ll,ll> > q;
	f(i,0,n-1){
		cin>>a[i];
		q.push({a[i],1});
		
	}
	
	while(!q.empty()){
		auto it=q.front();
		q.pop();
	//	cout<<it.F<<" "<<it.S<<'\n';
		if(it.F%x==0){
			sum+=(it.F)*it.S;
			q.push({it.F/x,it.S*x});
		}
		else{
			sum+=(it.F)*it.S;
			break;
		}
	}
//	cout<<sum<<'\n';
	while(!q.empty()){
		auto it=q.front();
	//	cout<<it.F<<" "<<it.S<<'\n';
		sum+=(it.F)*it.S;
		q.pop();
	}
	cout<<sum<<'\n';
}
int main()
{	Fast;
	int t;cin>>t;
	while(t--)
	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
// Problem: B. Strange List
// Contest: Codeforces - Codeforces Round #694 (Div. 2)
// URL: https://codeforces.com/contest/1471/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms

//code by: Ravi Khatri @Ravi_Orz
#include<bits/stdc++.h>
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define F first
#define S second
#define pb push_back
#define f(i,s,n) for(int i=(int)s;i<=(int)n;++i)
#define fr(i,n,s) for(int i=(int)n;i>=s;--i)
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define Fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;
#define mod 1000000007
ll n,m,l,r,x;
ll a[1000006];
ll b[1000006];
vi g[1000006];

void solve()
{	ll sum=0;
	cin>>n>>x;
	queue< pair<ll,ll> > q;
	f(i,0,n-1){
		cin>>a[i];
		q.push({a[i],1});
		
	}
	
	while(!q.empty()){
		auto it=q.front();
		q.pop();
	//	cout<<it.F<<" "<<it.S<<'\n';
		if(it.F%x==0){
			sum+=(it.F)*it.S;
			q.push({it.F/x,it.S*x});
		}
		else{
			sum+=(it.F)*it.S;
			break;
		}
	}
//	cout<<sum<<'\n';
	while(!q.empty()){
		auto it=q.front();
	//	cout<<it.F<<" "<<it.S<<'\n';
		sum+=(it.F)*it.S;
		q.pop();
	}
	cout<<sum<<'\n';
}
int main()
{	Fast;
	int t;cin>>t;
	while(t--)
	solve();
}
```