#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int n;
ll a[100005],x;

ll power(int x,int y){
    ll ans=1;
    while(y>0){
        if(y&1){
            ans = ans*x;
        }
        y = y/2;
        x = x*x;
    }
    return ans;
}

void solve(){
    cin >> n >>x;
    ll sum=0;
    for(int i=0;i<n;i++){
        cin >> a[i];
        sum += a[i];
    }
    ll cnt=0;
    int flag=1,i=0;

    while(flag==1){
        ll p = power(x,cnt);
        if((a[i]/p)%x==0) sum += a[i++];
        else {
            flag=0;
            break;
        }

        if(i==n){
            i=0;
            cnt++;
        } 
    }
    cout << sum << "\n";
}

int main() {
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

### 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>
#define ll long long int
using namespace std;
int n;
ll a[100005],x;

ll power(int x,int y){
    ll ans=1;
    while(y>0){
        if(y&1){
            ans = ans*x;
        }
        y = y/2;
        x = x*x;
    }
    return ans;
}

void solve(){
    cin >> n >>x;
    ll sum=0;
    for(int i=0;i<n;i++){
        cin >> a[i];
        sum += a[i];
    }
    ll cnt=0;
    int flag=1,i=0;

    while(flag==1){
        ll p = power(x,cnt);
        if((a[i]/p)%x==0) sum += a[i++];
        else {
            flag=0;
            break;
        }

        if(i==n){
            i=0;
            cnt++;
        } 
    }
    cout << sum << "\n";
}

int main() {
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

```