/*
 * author :Sadik Hassan(_sad_)
 *
*/
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#include "bits/stdc++.h"
using namespace std;
using   ll =          long long;
using   ld =          long double;
#define nl            "\n"
#define pb            push_back
#define fi            first
#define se            second
#define MP            make_pair
#define PI            (acos(-1.0))
#define rep1(i,n)     for(int i=1;i<=n;i++)
#define rep(i,n)      for(int i=0;i<n;i++)
#define urep(i,n)     for(int i=n-1;i>=0;i--)
#define urep1(i,n)    for(int i=n;i>=1;i--)
#define SZ(s)         (int)s.size()
#define all(x)        (x).begin(), (x).end()
#define rall(x)       (x).rbegin(), (x).rend()
#define w(t)          int t;cin>>t;while(t--)
#define _SAD()     	  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0),cout<<fixed<<setprecision(15);

typedef vector<int> vi;
typedef vector<ll> vii;
typedef set<int> si;
typedef set<ll> sii;
/*---------------------------------------------------------------------*/

const int N = (int)2e5+5;
const int MOD =(int)1e9+7;
const ll INF = (ll)1e18+5;

int main()
{
    _SAD()
    w(t)
    {
        int n,m;
        cin>>n>>m;
        ll c=0;
        vector<pair<ll,ll>>a;
        rep(i,n)
        {
            ll x;
            cin>>x;
            a.pb(MP(x,1));
        }
        rep(i,SZ(a))
        {
            if(a[i].fi%m!=0)break;
            else
            {
                ll y=a[i].fi/m;
                a.pb(MP(y,a[i].se*m));
            }
        }
        rep(i,SZ(a))
        {
            c+=(a[i].fi*a[i].se);
        }

        cout<<c<<nl;

    }
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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
/*
 * author :Sadik Hassan(_sad_)
 *
*/
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#include "bits/stdc++.h"
using namespace std;
using   ll =          long long;
using   ld =          long double;
#define nl            "\n"
#define pb            push_back
#define fi            first
#define se            second
#define MP            make_pair
#define PI            (acos(-1.0))
#define rep1(i,n)     for(int i=1;i<=n;i++)
#define rep(i,n)      for(int i=0;i<n;i++)
#define urep(i,n)     for(int i=n-1;i>=0;i--)
#define urep1(i,n)    for(int i=n;i>=1;i--)
#define SZ(s)         (int)s.size()
#define all(x)        (x).begin(), (x).end()
#define rall(x)       (x).rbegin(), (x).rend()
#define w(t)          int t;cin>>t;while(t--)
#define _SAD()     	  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0),cout<<fixed<<setprecision(15);

typedef vector<int> vi;
typedef vector<ll> vii;
typedef set<int> si;
typedef set<ll> sii;
/*---------------------------------------------------------------------*/

const int N = (int)2e5+5;
const int MOD =(int)1e9+7;
const ll INF = (ll)1e18+5;

int main()
{
    _SAD()
    w(t)
    {
        int n,m;
        cin>>n>>m;
        ll c=0;
        vector<pair<ll,ll>>a;
        rep(i,n)
        {
            ll x;
            cin>>x;
            a.pb(MP(x,1));
        }
        rep(i,SZ(a))
        {
            if(a[i].fi%m!=0)break;
            else
            {
                ll y=a[i].fi/m;
                a.pb(MP(y,a[i].se*m));
            }
        }
        rep(i,SZ(a))
        {
            c+=(a[i].fi*a[i].se);
        }

        cout<<c<<nl;

    }
    return 0;
}

```