#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define ff first
#define ss second
#define lli long long int
#define ll long long
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vpii vector<pii>
#define vpll vector<pll>
#define takeinp(arr,n) for(long long int i=0;i<n;i++) cin>>arr[i];
#define mii unordered_map<int,int>
#define mll unordered_map<ll,ll>
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repo(i,a,b) for(int i=a;i>b;i--)

            //GL HF//

int main(){
    int t;
    cin>>t;
    while(t--){
        ll i,j,n,x,sum=0;
        cin>>n>>x;
        ll arr[n],d[n];
        rep(i,0,n){
            cin>>arr[i];
            d[i]=arr[i];
        }
        ll min=INT_MAX;
        ll cpy[n],id;
        rep(i,0,n){
            ll c=1;
            while(d[i]%x==0){
                c++;
                d[i]/=x;
            }
            cpy[i]=c;
            if(c<min) {min=c;id=i;}
        }
        // cout<<min<<" "<<id<<endl;
        // for(auto a:cpy) cout<<a<<" ";
        // cout<<endl;
        rep(i,0,id){
            sum+=arr[i]*(min+1);
            // cout<<sum<<" ";
        }
        rep(i,id,n){
            sum+=arr[i]*min;
            // cout<<sum<<" ";
        }
        cout<<sum<<"\n";
    }   
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 pb push_back
#define ff first
#define ss second
#define lli long long int
#define ll long long
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vpii vector<pii>
#define vpll vector<pll>
#define takeinp(arr,n) for(long long int i=0;i<n;i++) cin>>arr[i];
#define mii unordered_map<int,int>
#define mll unordered_map<ll,ll>
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repo(i,a,b) for(int i=a;i>b;i--)

            //GL HF//

int main(){
    int t;
    cin>>t;
    while(t--){
        ll i,j,n,x,sum=0;
        cin>>n>>x;
        ll arr[n],d[n];
        rep(i,0,n){
            cin>>arr[i];
            d[i]=arr[i];
        }
        ll min=INT_MAX;
        ll cpy[n],id;
        rep(i,0,n){
            ll c=1;
            while(d[i]%x==0){
                c++;
                d[i]/=x;
            }
            cpy[i]=c;
            if(c<min) {min=c;id=i;}
        }
        // cout<<min<<" "<<id<<endl;
        // for(auto a:cpy) cout<<a<<" ";
        // cout<<endl;
        rep(i,0,id){
            sum+=arr[i]*(min+1);
            // cout<<sum<<" ";
        }
        rep(i,id,n){
            sum+=arr[i]*min;
            // cout<<sum<<" ";
        }
        cout<<sum<<"\n";
    }   
    return 0;
}

```