#include<bits/stdc++.h>
#include<stdlib.h>
#define pb push_back
#define mp make_pair
using namespace std ;
#define fi first
#define se second
#define ll long long
#include <sstream>  // for string streams
#include <string.h>
#include<iostream>
#define f(i,p,n) for(int i=p;i<n;i++)
#define fn(i,n,p) for(int i=n;i>=p;i--)
#define pf push_front
#define um unordered_map<ll,ll>  
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

bool sortbysec(const pair<pair<int,int>,int> &a, 
              const pair<pair<int,int>,int> &b) 
{ 
    if(a.fi!=b.fi)
    return a.fi.fi<b.fi.fi ;
    else
    return (a.fi.se<b.fi.se); 
}

int main() {
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t ;
    cin>>t ;
    while(t--){
        ll n,x ;
        cin>>n>>x ;
        ll flag=0 ;ll h[n+1] ;
        vector<ll> v[n+1] ;
        ll w=INT64_MAX,p ;
        f(i,1,n+1){
            ll a ;
            cin>>a ;h[i]=a ;
            v[i].pb(a) ;
            //if(a%x) flag=1 ;
            while(a%x==0&&flag==0){
                v[i].pb(a/x) ;
                a=a/x ;
            }
            if(v[i].size()<w){
                p= i ; w=v[i].size() ;
            }
        }
        ll s=0 ;//cout<<w<<"\n" ;
        f(i,1,p){
            ll k=v[i].size() ;
            f(j,0,min(w+1,k)){
                s+=h[i] ;
            }
        }
        f(i,p,n+1){
            ll k=v[i].size() ;
            f(j,0,min(w,k)) s+=h[i] ;
        }
        cout<<s<<"\n" ;
        
    }
    
               
}

### Prompt

Create a solution in cpp for 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>
#include<stdlib.h>
#define pb push_back
#define mp make_pair
using namespace std ;
#define fi first
#define se second
#define ll long long
#include <sstream>  // for string streams
#include <string.h>
#include<iostream>
#define f(i,p,n) for(int i=p;i<n;i++)
#define fn(i,n,p) for(int i=n;i>=p;i--)
#define pf push_front
#define um unordered_map<ll,ll>  
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

bool sortbysec(const pair<pair<int,int>,int> &a, 
              const pair<pair<int,int>,int> &b) 
{ 
    if(a.fi!=b.fi)
    return a.fi.fi<b.fi.fi ;
    else
    return (a.fi.se<b.fi.se); 
}

int main() {
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t ;
    cin>>t ;
    while(t--){
        ll n,x ;
        cin>>n>>x ;
        ll flag=0 ;ll h[n+1] ;
        vector<ll> v[n+1] ;
        ll w=INT64_MAX,p ;
        f(i,1,n+1){
            ll a ;
            cin>>a ;h[i]=a ;
            v[i].pb(a) ;
            //if(a%x) flag=1 ;
            while(a%x==0&&flag==0){
                v[i].pb(a/x) ;
                a=a/x ;
            }
            if(v[i].size()<w){
                p= i ; w=v[i].size() ;
            }
        }
        ll s=0 ;//cout<<w<<"\n" ;
        f(i,1,p){
            ll k=v[i].size() ;
            f(j,0,min(w+1,k)){
                s+=h[i] ;
            }
        }
        f(i,p,n+1){
            ll k=v[i].size() ;
            f(j,0,min(w,k)) s+=h[i] ;
        }
        cout<<s<<"\n" ;
        
    }
    
               
}

```