#include<stdio.h>
#include<iostream>
using namespace std;
typedef long long ll;
int main(){
	ll t;
	ll a[900001],cnt[900001];
	scanf("%d",&t);
	while(t--){
		ll n,x;
		scanf("%lld %lld",&n,&x);
		for(int i=0;i<n;i++) scanf("%lld",&a[i]);
		ll ans=0,s=1000000,num=0;
		for(int i=0;i<n;i++){
			ll y=0,p=a[i];
			while(p%x==0)
				y++,p=p/x;
			if(s>y) s=y,num=i;
		}
		//cout<<s<<"   "<<num<<"\n";
		for(int i=0;i<num;i++){
			ans+=(s+2)*a[i];
		}
		for(int i=num;i<n;i++)ans+=(s+1)*a[i];
		printf("%lld\n",ans);
	}
	return 0;
} 
		   	 		 		 	 			   	 	 				
### Prompt

Your challenge is to write 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<stdio.h>
#include<iostream>
using namespace std;
typedef long long ll;
int main(){
	ll t;
	ll a[900001],cnt[900001];
	scanf("%d",&t);
	while(t--){
		ll n,x;
		scanf("%lld %lld",&n,&x);
		for(int i=0;i<n;i++) scanf("%lld",&a[i]);
		ll ans=0,s=1000000,num=0;
		for(int i=0;i<n;i++){
			ll y=0,p=a[i];
			while(p%x==0)
				y++,p=p/x;
			if(s>y) s=y,num=i;
		}
		//cout<<s<<"   "<<num<<"\n";
		for(int i=0;i<num;i++){
			ans+=(s+2)*a[i];
		}
		for(int i=num;i<n;i++)ans+=(s+1)*a[i];
		printf("%lld\n",ans);
	}
	return 0;
} 
		   	 		 		 	 			   	 	 				
```