  #include<bits/stdc++.h>
  using namespace std;
  #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #define ll  long long
  #define ld long double
  #define pb push_back
  #define mp3 make_pair
  #define rep1(i,a,b) for(ll i=a;i<=b;i++)
  #define rep2(i,a,b) for(ll i=a;i<b;i++)
  const ll  mod=998244353;
  const ll INF=1e18;
  //template<typename T> T pow(T a,T b, long long m){T ans=1; while(b>0){ if(b%2==1) ans=(ans*a)%m; b/=2; a=(a*a)%m; } return ans%m; }
  template<typename T> T add(T a,T b,ll m){a+=b;while(a>=0)a-=m;while(a<0)a+=m; return a;}
  template<typename T> T mul(T a,T b,ll m){return (a*1ll*b)%m;}

 int main(){
   IOS;
   ll test;
   cin>>test;
   label: while(test--){
   ll n,x;
   cin>>n>>x;
   ll a[n];
   vector<pair<ll,ll>> v;
   ll sum=0;
   for(ll i=0;i<n;i++){
     cin>>a[i];
     v.pb({a[i],1});
   }
     for(ll i=0;i<v.size();i++){
      //  cout<<v[i].first<<endl;
       if(v[i].first%x==0){
         v.pb({v[i].first/x,x*v[i].second});
       }
       else 
       break;
     }
     for(auto it:v){
       sum+=it.first*it.second;
     }
     cout<<sum<<endl;
   }
   }
### Prompt

Please create a solution in CPP 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 IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #define ll  long long
  #define ld long double
  #define pb push_back
  #define mp3 make_pair
  #define rep1(i,a,b) for(ll i=a;i<=b;i++)
  #define rep2(i,a,b) for(ll i=a;i<b;i++)
  const ll  mod=998244353;
  const ll INF=1e18;
  //template<typename T> T pow(T a,T b, long long m){T ans=1; while(b>0){ if(b%2==1) ans=(ans*a)%m; b/=2; a=(a*a)%m; } return ans%m; }
  template<typename T> T add(T a,T b,ll m){a+=b;while(a>=0)a-=m;while(a<0)a+=m; return a;}
  template<typename T> T mul(T a,T b,ll m){return (a*1ll*b)%m;}

 int main(){
   IOS;
   ll test;
   cin>>test;
   label: while(test--){
   ll n,x;
   cin>>n>>x;
   ll a[n];
   vector<pair<ll,ll>> v;
   ll sum=0;
   for(ll i=0;i<n;i++){
     cin>>a[i];
     v.pb({a[i],1});
   }
     for(ll i=0;i<v.size();i++){
      //  cout<<v[i].first<<endl;
       if(v[i].first%x==0){
         v.pb({v[i].first/x,x*v[i].second});
       }
       else 
       break;
     }
     for(auto it:v){
       sum+=it.first*it.second;
     }
     cout<<sum<<endl;
   }
   }
```