#include<bits/stdc++.h>
#define ll long long int
#define vll vector<pair<ll, ll> >
#define vl vector<ll>
#define vi vector<int>
#define vii vector<pair<int, int> >
const ll mod = 1e9 + 7;
const ll inf = LLONG_MAX;
using namespace std;

ll power_mod(ll x, ll y, ll p)
{
	ll ans = 1;
	x = x%p;
	if(x==0)
	return p;

	while(y>0)
	{
		if(y&1)
		{
			ans = (ans*x)%p;
		}
		y = y>>1;
		x = (x*x)%p;
	}
	return ans;
} 

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int tt;
	cin>>tt;
	while(tt--)
	{
		ll i,j,k,l,m,n;
		ll x;
		cin>>n>>x;
		vl a(n, 0);
		
		
		ll sum=0;
		
		for(i=0;i<n;i++)
		{
		    cin>>a[i];
		    sum+=a[i];
		}
		ll maxdepth = inf;
		ll minpos = n;
		for(i=0;i<n;i++)
		{
		    
		    ll temp = 0;
		    ll temp2 = a[i];
		    while(temp2%x==0)
		    {
		        temp2/=x;
		        temp++;
		        if(temp2==0)
		        break;
		    }
		    if(temp<maxdepth)
		    {
		        maxdepth = temp;
		        minpos = i;
		    }
		    //cout<<"adding "<<temp*a[i]<<"\n";
		    //sum += temp*a[i];
		    
		}
		
		//cout<<"minpos:"<<minpos<<" maxdepth:"<<maxdepth<<"\n";
		
		for(i=0;i<n;i++)
		{
		    
		    ll temp = 0;
		    ll temp2 = a[i];
		    
		    if(i<minpos)
		    {
		        while(temp2%x==0)
    		    {
    		        temp2/=x;
    		        temp++;
    		        if(temp2==0)
    		        break;
    		    }
    		    temp = min(temp, maxdepth+1);
		    }
		    else
		    {
		        while(temp2%x==0)
    		    {
    		        temp2/=x;
    		        temp++;
    		        if(temp2==0)
    		        break;
    		    }
    		    temp = min(temp, maxdepth);
		    }
		    
		    
		    
		    //cout<<"adding "<<temp*a[i]<<"\n";
		    sum += temp*a[i];
		    
		}
		
		cout<<sum<<"\n";
	}
	return 0;
}

### Prompt

Generate 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 ll long long int
#define vll vector<pair<ll, ll> >
#define vl vector<ll>
#define vi vector<int>
#define vii vector<pair<int, int> >
const ll mod = 1e9 + 7;
const ll inf = LLONG_MAX;
using namespace std;

ll power_mod(ll x, ll y, ll p)
{
	ll ans = 1;
	x = x%p;
	if(x==0)
	return p;

	while(y>0)
	{
		if(y&1)
		{
			ans = (ans*x)%p;
		}
		y = y>>1;
		x = (x*x)%p;
	}
	return ans;
} 

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int tt;
	cin>>tt;
	while(tt--)
	{
		ll i,j,k,l,m,n;
		ll x;
		cin>>n>>x;
		vl a(n, 0);
		
		
		ll sum=0;
		
		for(i=0;i<n;i++)
		{
		    cin>>a[i];
		    sum+=a[i];
		}
		ll maxdepth = inf;
		ll minpos = n;
		for(i=0;i<n;i++)
		{
		    
		    ll temp = 0;
		    ll temp2 = a[i];
		    while(temp2%x==0)
		    {
		        temp2/=x;
		        temp++;
		        if(temp2==0)
		        break;
		    }
		    if(temp<maxdepth)
		    {
		        maxdepth = temp;
		        minpos = i;
		    }
		    //cout<<"adding "<<temp*a[i]<<"\n";
		    //sum += temp*a[i];
		    
		}
		
		//cout<<"minpos:"<<minpos<<" maxdepth:"<<maxdepth<<"\n";
		
		for(i=0;i<n;i++)
		{
		    
		    ll temp = 0;
		    ll temp2 = a[i];
		    
		    if(i<minpos)
		    {
		        while(temp2%x==0)
    		    {
    		        temp2/=x;
    		        temp++;
    		        if(temp2==0)
    		        break;
    		    }
    		    temp = min(temp, maxdepth+1);
		    }
		    else
		    {
		        while(temp2%x==0)
    		    {
    		        temp2/=x;
    		        temp++;
    		        if(temp2==0)
    		        break;
    		    }
    		    temp = min(temp, maxdepth);
		    }
		    
		    
		    
		    //cout<<"adding "<<temp*a[i]<<"\n";
		    sum += temp*a[i];
		    
		}
		
		cout<<sum<<"\n";
	}
	return 0;
}

```