#include <bits/stdc++.h>

#define ll                 long long
#define sf                 scanf
#define pf                 printf
#define nl                 printf("\n")
#define endl               '\n'
#define pb                 push_back
#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 TEST               int Test;cin>>Test;for(int _t=1;_t<=Test;_t++)
#define isValid(a,b,A,B)   (0<=a&&a<A) and (0<=b&&b<B)

using namespace std;

template <typename... T> void read(T &... args){ ((cin >> args), ...); }

void yes(){ pf("Yes\n"); }
void no() { pf("No\n"); }

int main()
{	
	ll n,x,a;
	TEST{
		sf("%lld %lld",&n,&x);
		vector<ll> vec;
		ll mx = 0;
		rep(i,0,n) sf("%lld",&a), mx=max(mx,a), vec.pb(a);
		ll sum = 0;
		rep(i,0,n) sum += vec[i];
		ll pow = x;
		while(pow<=mx){
			rep(i,0,n){
				if(vec[i]%pow==0) sum += vec[i];
				else{
					mx = -1;
					break;
				}
			}
			pow *= x;
		}
		pf("%lld\n",sum);
	}
	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>

#define ll                 long long
#define sf                 scanf
#define pf                 printf
#define nl                 printf("\n")
#define endl               '\n'
#define pb                 push_back
#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 TEST               int Test;cin>>Test;for(int _t=1;_t<=Test;_t++)
#define isValid(a,b,A,B)   (0<=a&&a<A) and (0<=b&&b<B)

using namespace std;

template <typename... T> void read(T &... args){ ((cin >> args), ...); }

void yes(){ pf("Yes\n"); }
void no() { pf("No\n"); }

int main()
{	
	ll n,x,a;
	TEST{
		sf("%lld %lld",&n,&x);
		vector<ll> vec;
		ll mx = 0;
		rep(i,0,n) sf("%lld",&a), mx=max(mx,a), vec.pb(a);
		ll sum = 0;
		rep(i,0,n) sum += vec[i];
		ll pow = x;
		while(pow<=mx){
			rep(i,0,n){
				if(vec[i]%pow==0) sum += vec[i];
				else{
					mx = -1;
					break;
				}
			}
			pow *= x;
		}
		pf("%lld\n",sum);
	}
	return 0;
}

```