#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 200010;
int tc, cs = 1;
int n;
ll x, arr[N], cnt[N];

ll getval(ll num, ll cnt) {
  ll ret = 0, p = 1;
  for (int step = 0; step <= cnt; ++step) {
    ret += p*num;
    num /= x;
    p *= x;
  }
  return ret;
}

void test_cases() {
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) cnt[i] = 0;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  ll mn = INT_MAX, mnidx = -1;
  for (int i = 1; i <= n; ++i) {
    ll tmp = arr[i];
    while (tmp % x == 0) {
      cnt[i]++;
      tmp /= x;
    }
    if (cnt[i] < mn) {
      mn = cnt[i];
      mnidx = i;
    }
  }
  ll ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (i < mnidx) ans += getval(arr[i], min(mn+1, cnt[i]));
    else ans += getval(arr[i], mn);
  }
  cout << ans << '\n';
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    test_cases();
  }
}
### 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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 200010;
int tc, cs = 1;
int n;
ll x, arr[N], cnt[N];

ll getval(ll num, ll cnt) {
  ll ret = 0, p = 1;
  for (int step = 0; step <= cnt; ++step) {
    ret += p*num;
    num /= x;
    p *= x;
  }
  return ret;
}

void test_cases() {
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) cnt[i] = 0;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  ll mn = INT_MAX, mnidx = -1;
  for (int i = 1; i <= n; ++i) {
    ll tmp = arr[i];
    while (tmp % x == 0) {
      cnt[i]++;
      tmp /= x;
    }
    if (cnt[i] < mn) {
      mn = cnt[i];
      mnidx = i;
    }
  }
  ll ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (i < mnidx) ans += getval(arr[i], min(mn+1, cnt[i]));
    else ans += getval(arr[i], mn);
  }
  cout << ans << '\n';
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    test_cases();
  }
}
```