#include <iostream>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <unordered_map>

using namespace std;

#define int int64_t
#define FOR(k,a,b) for(int k=(a); k<(b); ++k)
#define REP(k,a) for(int k=0; k<(a);++k)
#define SZ(a) int((a).size())
#define ALL(c) (c).begin(),(c).end()
#define MOD 1000000007

typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector< vector<int> > vvi;

signed main()
{
    ios_base::sync_with_stdio(false);

    #ifndef ONLINE_JUDGE
    freopen("sample2.txt", "r", stdin);
    #endif

    int T;
    cin >> T;
    REP(tc, T) {
        int N, X; 
        cin >> N >> X;
        int sum = 0, mini = 0;

        vi strange(N, 0);
        vi pf(N, 0);
        REP(i, N) {
            int xi;
            cin >> xi;
            strange[i] = xi;
            sum += xi;

            while (xi % X == 0) {
                xi /= X;
                pf[i]++;
            }

            if (i == 0)
                mini = pf[i];
            else if (mini > pf[i]) 
                mini = pf[i];
        }

        bool fin = false;
        REP(i, N) {
            if (pf[i] == mini) {
                fin = true;
            }

            if (fin) {
                sum += mini * strange[i];
            } else {
                sum += (mini + 1) * strange[i];
            }
        }

        cout << sum << endl;
    }

    return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 <queue>
#include <vector>
#include <set>
#include <stack>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <unordered_map>

using namespace std;

#define int int64_t
#define FOR(k,a,b) for(int k=(a); k<(b); ++k)
#define REP(k,a) for(int k=0; k<(a);++k)
#define SZ(a) int((a).size())
#define ALL(c) (c).begin(),(c).end()
#define MOD 1000000007

typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector< vector<int> > vvi;

signed main()
{
    ios_base::sync_with_stdio(false);

    #ifndef ONLINE_JUDGE
    freopen("sample2.txt", "r", stdin);
    #endif

    int T;
    cin >> T;
    REP(tc, T) {
        int N, X; 
        cin >> N >> X;
        int sum = 0, mini = 0;

        vi strange(N, 0);
        vi pf(N, 0);
        REP(i, N) {
            int xi;
            cin >> xi;
            strange[i] = xi;
            sum += xi;

            while (xi % X == 0) {
                xi /= X;
                pf[i]++;
            }

            if (i == 0)
                mini = pf[i];
            else if (mini > pf[i]) 
                mini = pf[i];
        }

        bool fin = false;
        REP(i, N) {
            if (pf[i] == mini) {
                fin = true;
            }

            if (fin) {
                sum += mini * strange[i];
            } else {
                sum += (mini + 1) * strange[i];
            }
        }

        cout << sum << endl;
    }

    return 0;
}

```