#include<bits/stdc++.h>
using namespace std;
#define Go_ ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long int
#define llu long long unsigned int
#define pii pair<int ,int>
#define PII pair<ll ,ll>

const int Max=1e9+100;
const ll  MAX=1e18+500;
const int sz=2e5+100;

ll sum=0;

void fun(vector<PII> test,ll x)
{
    bool bl=true;

    ll siz=test.size();
    vector<PII> le;

    for(auto xx: test)
    {
        if(xx.first % x != 0) {
            bl=false;
            break;
        }

        else{
            ll p=xx.first/x;
            sum+=(p * (xx.second * x));
            le.push_back({p,(xx.second * x)});
        }
    }

    if(bl) fun(le,x);
}

int main()
{
    Go_
    ll a,b,c,i,j,k,q,p,x,y,ct,ct1,m,l,r,x1,y1,mn,h,sum1,in,z,mid,n,mx;
    char ch;
    double d;
    string str1,str2,str;
    bool bl,bl1;
    int t,cs=1;

    cin>>t;
    while(t--)
    {
       cin>>n>>x;
       
       vector<PII> le;

       sum=0;
       for(int i=1;i<=n;i++)
       {
           cin>>p;
           le.push_back({p,1ll});
           sum+=p;
       }

       fun(le,x);
       cout<<sum<<'\n';
    }

   
}

### Prompt

Please provide a cpp coded solution 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>
using namespace std;
#define Go_ ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ll long long int
#define llu long long unsigned int
#define pii pair<int ,int>
#define PII pair<ll ,ll>

const int Max=1e9+100;
const ll  MAX=1e18+500;
const int sz=2e5+100;

ll sum=0;

void fun(vector<PII> test,ll x)
{
    bool bl=true;

    ll siz=test.size();
    vector<PII> le;

    for(auto xx: test)
    {
        if(xx.first % x != 0) {
            bl=false;
            break;
        }

        else{
            ll p=xx.first/x;
            sum+=(p * (xx.second * x));
            le.push_back({p,(xx.second * x)});
        }
    }

    if(bl) fun(le,x);
}

int main()
{
    Go_
    ll a,b,c,i,j,k,q,p,x,y,ct,ct1,m,l,r,x1,y1,mn,h,sum1,in,z,mid,n,mx;
    char ch;
    double d;
    string str1,str2,str;
    bool bl,bl1;
    int t,cs=1;

    cin>>t;
    while(t--)
    {
       cin>>n>>x;
       
       vector<PII> le;

       sum=0;
       for(int i=1;i<=n;i++)
       {
           cin>>p;
           le.push_back({p,1ll});
           sum+=p;
       }

       fun(le,x);
       cout<<sum<<'\n';
    }

   
}

```