#include <bits/stdc++.h>
using namespace std;    
#define int             long long
#define all(x)          x.begin(), x.end()
#define allr(x)         x.rbegin(), x.rend()
#define rep(i,a,b)      for(int i=a;i<b;i++)
#define repr(i,a,b)     for(int i=a;i>=b;i--)
#define endl            '\n'
#define pi              3.141592653589793238
#define mp              make_pair
#define pb              push_back
#define precise         cout<<fixed<<setprecision(20)
#define ff              first
#define ss              second
#define uniq(v)         v.resize(distance(v.begin(),unique(all(v))))
#define lb              lower_bound
#define ub              upper_bound
#define pii             pair<int,int>
#define vi              vector<int>
#define vp              vector<pair<int,int>>
#define vvi             vector<vector<int>>
#define deb(x)          cerr<<#x<<' '<<'='<<' '<<x<<'\n'
const int inf=1e18;
const int N=1000001;
int mod=1000000007;
//int mod=998244353;

int min(int a,int b){if(a<b) return a; else return b;}
int max(int a,int b){if(a>b) return a; else return b;}
int power(int a,int b){int res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int gcd(int p, int q){if(p%q==0) return q; else{return gcd(q,p%q);}} 
int mul(int a,int b){return ((a%mod)*(b%mod))%mod;}
int sub(int a,int b){return (((a-b)%mod)+mod)%mod;}

void solve() 
{
        int n,x;
        cin>>n>>x;
        int a[n];
        rep(i,0,n)
        cin>>a[i];
        int ans=0;
        vi b(n);
        rep(i,0,n)
        {
                int temp=a[i];
                int cnt=0;
                while(temp%x==0)
                {
                        temp/=x;
                        cnt++;
                }
                b[i]=cnt;
        }
        int m=min_element(all(b))-b.begin();
        rep(i,0,n)
        {
                if(i<m)
                {
                        ans+=(b[m]+2)*a[i];
                }
                else
                {
                        ans+=(b[m]+1)*a[i];
                }
        }
        cout<<ans<<endl;        
}

signed main()
{
        ios::sync_with_stdio(0); 
        cin.tie(0);cout.tie(0);
        int t=1;
        cin>>t;
        rep(i,0,t)
        {
                //cout<<"Case #"<<t+1<<": ";
                solve();
        }
        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 int             long long
#define all(x)          x.begin(), x.end()
#define allr(x)         x.rbegin(), x.rend()
#define rep(i,a,b)      for(int i=a;i<b;i++)
#define repr(i,a,b)     for(int i=a;i>=b;i--)
#define endl            '\n'
#define pi              3.141592653589793238
#define mp              make_pair
#define pb              push_back
#define precise         cout<<fixed<<setprecision(20)
#define ff              first
#define ss              second
#define uniq(v)         v.resize(distance(v.begin(),unique(all(v))))
#define lb              lower_bound
#define ub              upper_bound
#define pii             pair<int,int>
#define vi              vector<int>
#define vp              vector<pair<int,int>>
#define vvi             vector<vector<int>>
#define deb(x)          cerr<<#x<<' '<<'='<<' '<<x<<'\n'
const int inf=1e18;
const int N=1000001;
int mod=1000000007;
//int mod=998244353;

int min(int a,int b){if(a<b) return a; else return b;}
int max(int a,int b){if(a>b) return a; else return b;}
int power(int a,int b){int res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int gcd(int p, int q){if(p%q==0) return q; else{return gcd(q,p%q);}} 
int mul(int a,int b){return ((a%mod)*(b%mod))%mod;}
int sub(int a,int b){return (((a-b)%mod)+mod)%mod;}

void solve() 
{
        int n,x;
        cin>>n>>x;
        int a[n];
        rep(i,0,n)
        cin>>a[i];
        int ans=0;
        vi b(n);
        rep(i,0,n)
        {
                int temp=a[i];
                int cnt=0;
                while(temp%x==0)
                {
                        temp/=x;
                        cnt++;
                }
                b[i]=cnt;
        }
        int m=min_element(all(b))-b.begin();
        rep(i,0,n)
        {
                if(i<m)
                {
                        ans+=(b[m]+2)*a[i];
                }
                else
                {
                        ans+=(b[m]+1)*a[i];
                }
        }
        cout<<ans<<endl;        
}

signed main()
{
        ios::sync_with_stdio(0); 
        cin.tie(0);cout.tie(0);
        int t=1;
        cin>>t;
        rep(i,0,t)
        {
                //cout<<"Case #"<<t+1<<": ";
                solve();
        }
        return 0;
}
```