#include <bits/stdc++.h>

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

#define forn(i,n) for(int i = 0; i < int(n); i++)
#define forsn(i,s,n) for(int i = int(s); i < int(n); i++)
#define dforn(i,n) for (int i = int(n)-1; i >= 0; i--)
#define dforsn(i,s,n) for(int i = int(n)-1; i >= int(s); i--)
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define fst first
#define snd second
#define FAST_IO ios::sync_with_stdio(false);cin.tie(nullptr);

using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> ii;

int main() {
    FAST_IO;

    int t; cin >> t;
    forn(i,t) {
        int n,x; cin >> n >> x;
        ll s = 0;
        queue<pair<ll,ll>> arr;
        forn(j,n) {int aux; cin >> aux; arr.push({aux,1});}
        bool active = true;
        while (!arr.empty()) {
            pair<ll,ll> me = arr.front(); arr.pop();
            s += me.fst*(ll)me.snd;
            if (me.fst%x) active = false;
            if (active) arr.push({me.fst/x,(ll)x*me.snd});
        }
        cout << s << '\n';
    }

    return 0;
}

/// ¡¡¡¡¡ HACE CASOS DE PRUEBAAAAAAAAAAAAAAAA !!!!!!!!!
/// ESCRIBÍ en vez de tanto dar vueltas
/// si te parece que no va PROBALO PRIMERO!
/// CODEA LO BÁSICO PRIMERO!

### 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 <bits/stdc++.h>

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

#define forn(i,n) for(int i = 0; i < int(n); i++)
#define forsn(i,s,n) for(int i = int(s); i < int(n); i++)
#define dforn(i,n) for (int i = int(n)-1; i >= 0; i--)
#define dforsn(i,s,n) for(int i = int(n)-1; i >= int(s); i--)
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define fst first
#define snd second
#define FAST_IO ios::sync_with_stdio(false);cin.tie(nullptr);

using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> ii;

int main() {
    FAST_IO;

    int t; cin >> t;
    forn(i,t) {
        int n,x; cin >> n >> x;
        ll s = 0;
        queue<pair<ll,ll>> arr;
        forn(j,n) {int aux; cin >> aux; arr.push({aux,1});}
        bool active = true;
        while (!arr.empty()) {
            pair<ll,ll> me = arr.front(); arr.pop();
            s += me.fst*(ll)me.snd;
            if (me.fst%x) active = false;
            if (active) arr.push({me.fst/x,(ll)x*me.snd});
        }
        cout << s << '\n';
    }

    return 0;
}

/// ¡¡¡¡¡ HACE CASOS DE PRUEBAAAAAAAAAAAAAAAA !!!!!!!!!
/// ESCRIBÍ en vez de tanto dar vueltas
/// si te parece que no va PROBALO PRIMERO!
/// CODEA LO BÁSICO PRIMERO!

```