#include<bits/stdc++.h>
using namespace std;

int main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
#endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 0;
    cin >> t;
    while(t--) {
        int n = 0, x = 0;
        cin >> n >> x;

        vector<long long> v(n, 0);
        vector<int> tm(n, 0);
        for(int i = 0; i < n; i ++) cin >> v[i];

        long long sum = 0;
        
        
        for(int i = 0; i < n; i++) {
            int temp = v[i];
            int cnt = 0;
            while(temp % x == 0) {
                cnt++;
                temp = temp/x;
            }
            tm[i] = cnt;
        }

        long long mn = INT_MAX;
        for(int i = 0; i < n; i++) mn = min(mn, (long long)tm[i]);
        int found = 0;
        // for(int i = 0; i < n; i++) {
        //     sum += v[i];
        // }
        // for(int i = 0; i < n; i++) cout << tm[i] << " ";
        // cout << endl;
        for(int i = 0; i < n; i++) sum += v[i];
        for(int i = 0; i < n; i++) sum += (long long) (mn) * v[i];
        for(int i = 0; i < n; i++) {
            if(tm[i] != mn) {
                sum += v[i];
            }
            else break;
        }


        cout << sum << endl;
    }

    return 0;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;

int main() {

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
#endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 0;
    cin >> t;
    while(t--) {
        int n = 0, x = 0;
        cin >> n >> x;

        vector<long long> v(n, 0);
        vector<int> tm(n, 0);
        for(int i = 0; i < n; i ++) cin >> v[i];

        long long sum = 0;
        
        
        for(int i = 0; i < n; i++) {
            int temp = v[i];
            int cnt = 0;
            while(temp % x == 0) {
                cnt++;
                temp = temp/x;
            }
            tm[i] = cnt;
        }

        long long mn = INT_MAX;
        for(int i = 0; i < n; i++) mn = min(mn, (long long)tm[i]);
        int found = 0;
        // for(int i = 0; i < n; i++) {
        //     sum += v[i];
        // }
        // for(int i = 0; i < n; i++) cout << tm[i] << " ";
        // cout << endl;
        for(int i = 0; i < n; i++) sum += v[i];
        for(int i = 0; i < n; i++) sum += (long long) (mn) * v[i];
        for(int i = 0; i < n; i++) {
            if(tm[i] != mn) {
                sum += v[i];
            }
            else break;
        }


        cout << sum << endl;
    }

    return 0;
}
```