#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
#define sz(v) (int)v.size() 
#define all(v) v.begin(), v.end()
void dbg_out() { cerr << "\b\b]\n"; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T){ cerr << H << ", "; dbg_out(T...);}
#define watch(...) cerr << "[" << #__VA_ARGS__ << "]: [", dbg_out(__VA_ARGS__)

int bound(int n, int x) {
  int cnts = 0;
  while (n % x == 0) {
    cnts++;
    n /= x;
  }
  return cnts;
}

void solve() {
  int n, x; cin >> n >> x;
  int idx = -1, mn = 1e9;
  vector <ll> A(n);
  ll sum = 0;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    sum += A[i];
    int cnts = bound(A[i], x);
    if (cnts < mn) {
      mn = cnts;
      idx = i;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (i < idx) {
      sum += (A[i] * (mn + 1)); 
    }
    else {
      sum += (A[i] * mn);
    }
  }
  cout << sum << endl;
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr);

    int t; cin >> t;
    while (t--) solve();

    return 0;
}


### Prompt

Create a solution in Cpp for 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>
using namespace std;
#define ll long long
#define endl '\n'
#define sz(v) (int)v.size() 
#define all(v) v.begin(), v.end()
void dbg_out() { cerr << "\b\b]\n"; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T){ cerr << H << ", "; dbg_out(T...);}
#define watch(...) cerr << "[" << #__VA_ARGS__ << "]: [", dbg_out(__VA_ARGS__)

int bound(int n, int x) {
  int cnts = 0;
  while (n % x == 0) {
    cnts++;
    n /= x;
  }
  return cnts;
}

void solve() {
  int n, x; cin >> n >> x;
  int idx = -1, mn = 1e9;
  vector <ll> A(n);
  ll sum = 0;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    sum += A[i];
    int cnts = bound(A[i], x);
    if (cnts < mn) {
      mn = cnts;
      idx = i;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (i < idx) {
      sum += (A[i] * (mn + 1)); 
    }
    else {
      sum += (A[i] * mn);
    }
  }
  cout << sum << endl;
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr);

    int t; cin >> t;
    while (t--) solve();

    return 0;
}


```