#include <bits/stdc++.h>
#define arr(i,n) for(ll i=0;i<n;i++)
#define ARR(i,n) for(ll i=n-1;i>=0;i--)
#define f(i,a,b)  for(ll i=a;i<b;i++)
#define F(i,a,b)  for(ll i=b-1;i>=a;i--)
#define float double
#define PI 3.14159265358979323846
#define pb push_back
#define mp make_pair
#define vi vector<ll>
#define ceil(a,b) ceil(a*1.0/b)
#define vpii vector< pair<ll,ll> >
#define debug(x) cout<<(#x)<<" is "<<(x)<<endl
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define all(v) v.begin(),v.end()
#define inputarr(a,n) for(ll i=0;i<n;++i) cin>>a[i]
#define integer_to_string to_string 
#define IOS  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
typedef long long ll;
using namespace std;
const ll  mod = 1000000000 + 7;
void solve()
{
    //code start here
    ll n,x;
    cin>>n>>x;
    ll mini=INT_MAX;
    ll a[n],id=0,res=0;
    arr(i,n)
    {
        cin>>a[i];
        res+=a[i];
        ll temp=a[i],cnt=0;
        while(temp%x==0) {cnt++; temp/=x;}
        if(mini>cnt) {id=i; mini=cnt;}
    }
    ll ans=(mini+1)*res;
    res=0;
    arr(i,id) res+=a[i]; 
    cout<<ans+res<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
      solve();
    return 0;
}
### Prompt

Please formulate 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>
#define arr(i,n) for(ll i=0;i<n;i++)
#define ARR(i,n) for(ll i=n-1;i>=0;i--)
#define f(i,a,b)  for(ll i=a;i<b;i++)
#define F(i,a,b)  for(ll i=b-1;i>=a;i--)
#define float double
#define PI 3.14159265358979323846
#define pb push_back
#define mp make_pair
#define vi vector<ll>
#define ceil(a,b) ceil(a*1.0/b)
#define vpii vector< pair<ll,ll> >
#define debug(x) cout<<(#x)<<" is "<<(x)<<endl
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define all(v) v.begin(),v.end()
#define inputarr(a,n) for(ll i=0;i<n;++i) cin>>a[i]
#define integer_to_string to_string 
#define IOS  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
typedef long long ll;
using namespace std;
const ll  mod = 1000000000 + 7;
void solve()
{
    //code start here
    ll n,x;
    cin>>n>>x;
    ll mini=INT_MAX;
    ll a[n],id=0,res=0;
    arr(i,n)
    {
        cin>>a[i];
        res+=a[i];
        ll temp=a[i],cnt=0;
        while(temp%x==0) {cnt++; temp/=x;}
        if(mini>cnt) {id=i; mini=cnt;}
    }
    ll ans=(mini+1)*res;
    res=0;
    arr(i,id) res+=a[i]; 
    cout<<ans+res<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
      solve();
    return 0;
}
```