#include <bits/stdc++.h>
#define s second
#define f first
#define pb push_back
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define _ ios_base::sync_with_stdio(0);cin.tie(0);
#define read(x) for(auto& qw : (x)) cin >> qw;
#define print(x) for(auto& qw : (x)) cout << qw << " "; cout << endl;
using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;

ll fexp(ll a, ll e){
	ll ans = 1;
	while(e){
		if(e & 1)
			ans *= a;
		a *= a;
		e >>= 1;
	}
	return ans;
}

int N;
ll K;
vector<int> V;

ll solve(ll mul){

	bool ok = 1;
	ll sum = 0;
	for(int i = 0; i < N; i++){
		ok &= !(V[i] % K);

		V[i] /= K;
		if(ok)
			sum += V[i]*mul;// cout << "adding " << V[i] << " " << V[i]*mul << endl;
		else break;
	}

	return sum + (ok ? solve(mul*K) : 0);
}

void test_case(){
	cin >> N >> K;

	V.resize(N);
	read(V);

	ll sum = 0;
	for(int i : V) sum += i;
	ll ans = solve(K);

	cout << ans + sum << endl;
}

int main(){ 

	int tt = 1;
	cin >> tt;

	while(tt--){
		test_case();
	}

	return 0;
}

### Prompt

In Cpp, your task is to solve 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 s second
#define f first
#define pb push_back
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define _ ios_base::sync_with_stdio(0);cin.tie(0);
#define read(x) for(auto& qw : (x)) cin >> qw;
#define print(x) for(auto& qw : (x)) cout << qw << " "; cout << endl;
using namespace std;

typedef long long ll;
typedef pair<int,int> ii;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;

ll fexp(ll a, ll e){
	ll ans = 1;
	while(e){
		if(e & 1)
			ans *= a;
		a *= a;
		e >>= 1;
	}
	return ans;
}

int N;
ll K;
vector<int> V;

ll solve(ll mul){

	bool ok = 1;
	ll sum = 0;
	for(int i = 0; i < N; i++){
		ok &= !(V[i] % K);

		V[i] /= K;
		if(ok)
			sum += V[i]*mul;// cout << "adding " << V[i] << " " << V[i]*mul << endl;
		else break;
	}

	return sum + (ok ? solve(mul*K) : 0);
}

void test_case(){
	cin >> N >> K;

	V.resize(N);
	read(V);

	ll sum = 0;
	for(int i : V) sum += i;
	ll ans = solve(K);

	cout << ans + sum << endl;
}

int main(){ 

	int tt = 1;
	cin >> tt;

	while(tt--){
		test_case();
	}

	return 0;
}

```