#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fo(i,n) for(i=0;i<n;i++)
#define forev(i,n) for(i=n-1;i>=0;i--)
#define dfo(a,b,c) for(i=a;i<b;i+=c)
#define si(x)  scanf("%d", &x)
#define sl(x)  scanf("%lld", &x)
#define ss(s)  sacnf("%s", s)
#define pi(x)  printf("%d\n", x)
#define pl(x)  printf("%lld\n", x)
#define ps(s)  printf("%s\n", s)
#define deb(x) cout<< #x << "=" << x << endl;
#define deb2(x, y) cout<< #x << "=" << x <<","<< #y << "=" << y <<endl;
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define find(x,n) binary_search(all(x), n)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)

typedef pair<int, int> pii;
typedef pair<ll, ll>   pl; 
typedef vector<int>   vi;
typedef vector<ll>    vl;
typedef vector<vi>    vvi;
typedef vector<vl>    vvl;
typedef set<int>   si;
typedef set<ll>   sl;
typedef unordered_map<int, int> umi;
typedef map<int, int> mi;
typedef vector<pl>   vpl;
typedef queue<int>   qi;
typedef queue<ll>   ql;
const int MOD = 1000000007;   //(10^9 + 7)

/*
bool isprime(ll n) 
{ 
  if (n <= 1) return false; 
  if (n <= 3) return true; 

  if (n % 2 == 0 || n % 3 == 0) return false; 

  for (ll i = 5; i * i <= n; i += 6) 
	if (n % i == 0 || n % (i+2) == 0) 
	  return false; 

  return true; 
} 

bool prime[15000105]; 
void sieve(int n) 
{ 
  for (ll i = 0; i <= n; i++) prime[i] = 1;
  for (ll p = 2; p * p <= n; p++) 
  { 
	if (prime[p] == true) 
	{ 
	  for (ll i = p * p; i <= n; i += p) 
		prime[i] = false; 
	} 
  } 
  prime[1] = prime[0] = 0;
}

*/

ll gcdll(ll a,ll b) {
    if (b==0) return a;
    return gcdll(b,a%b);
}


int gcd(int a,int b) {
    if (b==0) return a;
    return gcd(b,a%b);
}



int mpow(int base, int exp)
{
	base %= MOD;
	int result = 1;
	while(exp>0)
	{
		if(exp & 1)
		{
			result = ((ll)result*base);
		}
		
		base = ((ll)base*base) % MOD;
		exp >>= 1;
	}
	
	return result;
}



void solve()
{
	ll i,j,n,m,a,b,rem,count=0,tot=0;
    bool temp = true;
    ql que;
    cin>>n>>a;
    vpl vp(n,{0,0});
    fo(i,n)
    {
        int ele;
        cin>>ele;
        vp[i].S=0;
        vp[i].F=ele;
        
        
    }

    while(temp)
    {
        fo(i,n)
        {
            if(vp[i].F%a)
            {
                temp= false;
                j=0;
                while(j<n)
                {
                    tot+=pow(a,vp[j].S)*vp[j].F;
                    j++;
                }
                break;
            }
            else
            {
                tot+=pow(a,vp[i].S)*vp[i].F;
                vp[i].F/=a;
                vp[i].S++;
                
            }
            
        }
    }
    cout<<tot<<endl;
}


int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		solve();
	}
	
	return 0;
}



### 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 ll long long
#define fo(i,n) for(i=0;i<n;i++)
#define forev(i,n) for(i=n-1;i>=0;i--)
#define dfo(a,b,c) for(i=a;i<b;i+=c)
#define si(x)  scanf("%d", &x)
#define sl(x)  scanf("%lld", &x)
#define ss(s)  sacnf("%s", s)
#define pi(x)  printf("%d\n", x)
#define pl(x)  printf("%lld\n", x)
#define ps(s)  printf("%s\n", s)
#define deb(x) cout<< #x << "=" << x << endl;
#define deb2(x, y) cout<< #x << "=" << x <<","<< #y << "=" << y <<endl;
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define find(x,n) binary_search(all(x), n)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)

typedef pair<int, int> pii;
typedef pair<ll, ll>   pl; 
typedef vector<int>   vi;
typedef vector<ll>    vl;
typedef vector<vi>    vvi;
typedef vector<vl>    vvl;
typedef set<int>   si;
typedef set<ll>   sl;
typedef unordered_map<int, int> umi;
typedef map<int, int> mi;
typedef vector<pl>   vpl;
typedef queue<int>   qi;
typedef queue<ll>   ql;
const int MOD = 1000000007;   //(10^9 + 7)

/*
bool isprime(ll n) 
{ 
  if (n <= 1) return false; 
  if (n <= 3) return true; 

  if (n % 2 == 0 || n % 3 == 0) return false; 

  for (ll i = 5; i * i <= n; i += 6) 
	if (n % i == 0 || n % (i+2) == 0) 
	  return false; 

  return true; 
} 

bool prime[15000105]; 
void sieve(int n) 
{ 
  for (ll i = 0; i <= n; i++) prime[i] = 1;
  for (ll p = 2; p * p <= n; p++) 
  { 
	if (prime[p] == true) 
	{ 
	  for (ll i = p * p; i <= n; i += p) 
		prime[i] = false; 
	} 
  } 
  prime[1] = prime[0] = 0;
}

*/

ll gcdll(ll a,ll b) {
    if (b==0) return a;
    return gcdll(b,a%b);
}


int gcd(int a,int b) {
    if (b==0) return a;
    return gcd(b,a%b);
}



int mpow(int base, int exp)
{
	base %= MOD;
	int result = 1;
	while(exp>0)
	{
		if(exp & 1)
		{
			result = ((ll)result*base);
		}
		
		base = ((ll)base*base) % MOD;
		exp >>= 1;
	}
	
	return result;
}



void solve()
{
	ll i,j,n,m,a,b,rem,count=0,tot=0;
    bool temp = true;
    ql que;
    cin>>n>>a;
    vpl vp(n,{0,0});
    fo(i,n)
    {
        int ele;
        cin>>ele;
        vp[i].S=0;
        vp[i].F=ele;
        
        
    }

    while(temp)
    {
        fo(i,n)
        {
            if(vp[i].F%a)
            {
                temp= false;
                j=0;
                while(j<n)
                {
                    tot+=pow(a,vp[j].S)*vp[j].F;
                    j++;
                }
                break;
            }
            else
            {
                tot+=pow(a,vp[i].S)*vp[i].F;
                vp[i].F/=a;
                vp[i].S++;
                
            }
            
        }
    }
    cout<<tot<<endl;
}


int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		solve();
	}
	
	return 0;
}



```