#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define maxN 200010


int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
    ll t;
    cin>>t;
    while(t--){
        ll n,x,sum=0;
        cin>>n>>x;
        vector<ll> v(n);
        vector<ll> p(n);
        for(ll i=0;i<n;i++){
            cin>>v[i];
            ll temp=v[i],cnt=0;
            while((temp%x)==0){
                cnt++;
                temp=temp/x;
            }
            sum+=v[i];
            p[i]=cnt;
        }
        // for(auto i: p){
        //     cout<< i<< " ";
        // }
        // cout<<endl;
        ll minimum= 1e10;
        for(ll i=0;i<n;i++){
            minimum=min(minimum,p[i]);
        }
        bool flag=false;
        sum+= (minimum*sum);
        for(int i=0;i<n;i++){
            if(p[i]==minimum){
                break;
            }else{
                sum+= v[i];
            }
        }
        cout<<sum<<endl;
    }
	return 0;
}
### Prompt

Your task is to create 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 <bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl "\n"
#define mod 1000000007
#define maxN 200010


int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
    ll t;
    cin>>t;
    while(t--){
        ll n,x,sum=0;
        cin>>n>>x;
        vector<ll> v(n);
        vector<ll> p(n);
        for(ll i=0;i<n;i++){
            cin>>v[i];
            ll temp=v[i],cnt=0;
            while((temp%x)==0){
                cnt++;
                temp=temp/x;
            }
            sum+=v[i];
            p[i]=cnt;
        }
        // for(auto i: p){
        //     cout<< i<< " ";
        // }
        // cout<<endl;
        ll minimum= 1e10;
        for(ll i=0;i<n;i++){
            minimum=min(minimum,p[i]);
        }
        bool flag=false;
        sum+= (minimum*sum);
        for(int i=0;i<n;i++){
            if(p[i]==minimum){
                break;
            }else{
                sum+= v[i];
            }
        }
        cout<<sum<<endl;
    }
	return 0;
}
```