#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#define endl '\n'
#define PI acos(-1)
// #define DEBUG
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int Mod = 1e9+7;
const int EXP = 1e-8;
// inline ll gcd(ll x, ll y){if(y==0) return x;return gcd(y,x%y);}//x>y
inline void debug(){printf("@@\n");}
const int M = 1e5+10;
ll n,x;
ll a[M];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifdef DEBUG
        freopen("C://Users//24887//Data-structures-and-Algorithms//input.in","r",stdin);
        freopen("C://Users//24887//Data-structures-and-Algorithms//output.out","w",stdout);
    #endif
    int t;
    cin >> t;
    while(t--){
        cin >> n >> x;
        int Min = INF;
        int point = 0;
        ll sum = 0;
        for(int i=0;i<n;i++){
            cin >> a[i];
            ll y = a[i];
            int cnt = 1;
            while(y%x==0){
                cnt++;
                y /= x;
            }
            if(Min>cnt){
                Min = cnt;
                point = i;
            }
        }
        for(int i=0;i<n;i++){
            if(i<point){
                sum += (a[i]*(Min+1));
            }else{
                sum += (a[i]*(Min));
            }
        }
        cout << sum << endl;
    }

    return 0;
}
### Prompt

Your task is to create 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 <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#define endl '\n'
#define PI acos(-1)
// #define DEBUG
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const int Mod = 1e9+7;
const int EXP = 1e-8;
// inline ll gcd(ll x, ll y){if(y==0) return x;return gcd(y,x%y);}//x>y
inline void debug(){printf("@@\n");}
const int M = 1e5+10;
ll n,x;
ll a[M];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    #ifdef DEBUG
        freopen("C://Users//24887//Data-structures-and-Algorithms//input.in","r",stdin);
        freopen("C://Users//24887//Data-structures-and-Algorithms//output.out","w",stdout);
    #endif
    int t;
    cin >> t;
    while(t--){
        cin >> n >> x;
        int Min = INF;
        int point = 0;
        ll sum = 0;
        for(int i=0;i<n;i++){
            cin >> a[i];
            ll y = a[i];
            int cnt = 1;
            while(y%x==0){
                cnt++;
                y /= x;
            }
            if(Min>cnt){
                Min = cnt;
                point = i;
            }
        }
        for(int i=0;i<n;i++){
            if(i<point){
                sum += (a[i]*(Min+1));
            }else{
                sum += (a[i]*(Min));
            }
        }
        cout << sum << endl;
    }

    return 0;
}
```