#include <bits/stdc++.h>
using namespace std;
#define lli long long int 
#define ll long long
#define vi vector<int>
#define vlli vector<lli>
#define si set<int>
#define vpii vector<pair<int, int>>
#define vsi vector<set<int>>
#define pb push_back 
void debug(int _a) {cout << _a << " ";}
template<typename T1, typename T2> void debug(pair<T1, T2> _p) {cout<<"{"<<_p.first<<": "<<_p.second<<"}"<< endl;}
template<typename T> void debug(vector<T> _aa) {for (auto i: _aa) debug(i); cout << endl;}
template<typename T> void debug(set<T> _aa) {for (auto i: _aa) debug(i); cout << endl;}
template<typename T1, typename T2> void debug(map<T1, T2> _mm) {for (auto i: _mm) debug(i);}

const lli INF = 1e9-1, MOD = 998244353;
const int N = 1e5; 
const int di[] = {-1,0,1,0}, dj[] = {0,1,0,-1};

void solve () {
    lli n, x; cin >> n >> x;
    queue<pair<lli, lli>> q;
    for (int i = 0; i < n; i++) {
        lli a; cin >> a;
        q.push({a, 1});
    }
    lli sum = 0;
    while (!q.empty()) {
        pair<lli, lli> cur = q.front(); q.pop();
        lli num = cur.first, cnt = cur.second;
        //debug(cur);
        sum += num * cnt;
        if (num % x == 0) {
            num = num/x;
            q.push({num, cnt*x});
        } else break;
    }
    while (!q.empty()) {
        pair<lli, lli> cur = q.front(); q.pop();
        //debug(cur);
        lli num = cur.first, cnt = cur.second;
        sum += num * cnt;
    }
    cout << sum << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    //cout << fixed << setprecision(9);
    int t = 1; cin >> t;
    while (t--) solve();
}
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;
#define lli long long int 
#define ll long long
#define vi vector<int>
#define vlli vector<lli>
#define si set<int>
#define vpii vector<pair<int, int>>
#define vsi vector<set<int>>
#define pb push_back 
void debug(int _a) {cout << _a << " ";}
template<typename T1, typename T2> void debug(pair<T1, T2> _p) {cout<<"{"<<_p.first<<": "<<_p.second<<"}"<< endl;}
template<typename T> void debug(vector<T> _aa) {for (auto i: _aa) debug(i); cout << endl;}
template<typename T> void debug(set<T> _aa) {for (auto i: _aa) debug(i); cout << endl;}
template<typename T1, typename T2> void debug(map<T1, T2> _mm) {for (auto i: _mm) debug(i);}

const lli INF = 1e9-1, MOD = 998244353;
const int N = 1e5; 
const int di[] = {-1,0,1,0}, dj[] = {0,1,0,-1};

void solve () {
    lli n, x; cin >> n >> x;
    queue<pair<lli, lli>> q;
    for (int i = 0; i < n; i++) {
        lli a; cin >> a;
        q.push({a, 1});
    }
    lli sum = 0;
    while (!q.empty()) {
        pair<lli, lli> cur = q.front(); q.pop();
        lli num = cur.first, cnt = cur.second;
        //debug(cur);
        sum += num * cnt;
        if (num % x == 0) {
            num = num/x;
            q.push({num, cnt*x});
        } else break;
    }
    while (!q.empty()) {
        pair<lli, lli> cur = q.front(); q.pop();
        //debug(cur);
        lli num = cur.first, cnt = cur.second;
        sum += num * cnt;
    }
    cout << sum << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    //cout << fixed << setprecision(9);
    int t = 1; cin >> t;
    while (t--) solve();
}
```