#include <bits/stdc++.h>

using namespace std;

int main(){
    //freopen("in", "r", stdin);
    int tc,n,x;
    cin>>tc;
    while(tc--){
        cin>>n>>x;
        vector<pair<int,long long>> vii;
        vector<int> v(n);
        for(int i=0;i<n;i++){
            cin>>v[i];
        }
        for(int i=0;i<n;){
            long long c=0,x=v[i];
            while(x==v[i] && i<n){
                c++;
                i++;
                if(i==n) break;
            }
            vii.push_back(make_pair(x,c));
        }
        for(int i=0;i<vii.size();i++){
            if(vii[i].first%x!=0){
                break;
            }else{
                vii.push_back(make_pair(vii[i].first/x, x*vii[i].second));
            }
        }
        long long s=0;
        for(int i=0;i<vii.size();i++){
            s+=vii[i].first*vii[i].second;
        }
        cout<<s<<endl;
    }
}

### 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>

using namespace std;

int main(){
    //freopen("in", "r", stdin);
    int tc,n,x;
    cin>>tc;
    while(tc--){
        cin>>n>>x;
        vector<pair<int,long long>> vii;
        vector<int> v(n);
        for(int i=0;i<n;i++){
            cin>>v[i];
        }
        for(int i=0;i<n;){
            long long c=0,x=v[i];
            while(x==v[i] && i<n){
                c++;
                i++;
                if(i==n) break;
            }
            vii.push_back(make_pair(x,c));
        }
        for(int i=0;i<vii.size();i++){
            if(vii[i].first%x!=0){
                break;
            }else{
                vii.push_back(make_pair(vii[i].first/x, x*vii[i].second));
            }
        }
        long long s=0;
        for(int i=0;i<vii.size();i++){
            s+=vii[i].first*vii[i].second;
        }
        cout<<s<<endl;
    }
}

```