#include <cmath>
#include <time.h>
#include <ctype.h>
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

#define sp system("pause")
#define FOR(i,a,b) for(int i=a;i<b;++i)
#define FORD(i,a,b) for(int i=a;i>=b;--i)
#define REP(i,n) FOR(i,0,(int)n-1)
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define MS0(x) memset(x,0,sizeof(x))
#define MS1(x) memset(x,1,sizeof(x))
#define SORT(a,n) sort(begin(a),begin(a)+n)
#define REV(a,n) reverse(begin(a),begin(a)+n)
#define ll long long
#define pii pair<int,int>
#define MOD 1000000007

const int mxN=2e5;

int main(){
    
    int t;
	cin>>t;
	while(t--)
	{
		ll sum=0;
		pair<int,int> a[mxN];
		int i,n,x;
		cin>>n>>x;
		for(i=0;i<n;i++)
		{
			cin>>a[i].first;
			a[i].second=a[i].first;
			sum+=a[i].first;
		}
		i=0;
		while(a[i].second%x==0)
		{
			sum+=a[i].first;
			a[i].second/=x;
			i++;
			i%=n;
		}
		cout<<sum<<endl;
	}
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <cmath>
#include <time.h>
#include <ctype.h>
#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

#define sp system("pause")
#define FOR(i,a,b) for(int i=a;i<b;++i)
#define FORD(i,a,b) for(int i=a;i>=b;--i)
#define REP(i,n) FOR(i,0,(int)n-1)
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define MS0(x) memset(x,0,sizeof(x))
#define MS1(x) memset(x,1,sizeof(x))
#define SORT(a,n) sort(begin(a),begin(a)+n)
#define REV(a,n) reverse(begin(a),begin(a)+n)
#define ll long long
#define pii pair<int,int>
#define MOD 1000000007

const int mxN=2e5;

int main(){
    
    int t;
	cin>>t;
	while(t--)
	{
		ll sum=0;
		pair<int,int> a[mxN];
		int i,n,x;
		cin>>n>>x;
		for(i=0;i<n;i++)
		{
			cin>>a[i].first;
			a[i].second=a[i].first;
			sum+=a[i].first;
		}
		i=0;
		while(a[i].second%x==0)
		{
			sum+=a[i].first;
			a[i].second/=x;
			i++;
			i%=n;
		}
		cout<<sum<<endl;
	}
	return 0;
}
```