#include <bits/stdc++.h>
using namespace std;
#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define ll  long long
#define int long long
#define ld  long double
#define endl "\n"
#define pb  push_back
#define fill(a,val) memset(a,val,sizeof(a))
#define ff  first
#define ss  second
#define test  ll t; cin>>t; while(t--)
#define loop(i,a,b)  for(ll i=a;i<b;i++)
#define loopr(i,a,b) for(ll i=a;i>=b;i--)
#define pii pair<ll,ll>
#define all(v) v.begin(),v.end()
istream& in(cin);
ostream& out(cout);
const ll mod     = 1000 * 1000 * 1000;
const ll inf     = 1ll*1000*1000*1000*1000*1000*1000 + 7;
const ll mod2    = 998244353;
const ll N       = 1000 * 1000 +100;
const ld pi      = 3.141592653589793;
ll power(ll x,ll y,ll p = LLONG_MAX ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}

void solve(){
    ll n,x;
    cin >> n >> x;
    vector<ll>a;
    vector<ll>fr;
    for(ll i=0;i<n;i++){
      ll y;
      cin >> y;
      a.pb(y);
      fr.pb(1);
    }
    ll ans=0,i;
    for(i=0;i<a.size();i++){
       if(a[i]%x==0){
         ll cnt=x*fr[i];
         a.pb(a[i]/x);
         fr.pb(cnt);
         ans+=fr[i]*a[i];
       }
       else{
         ans+=fr[i]*a[i];
         break;
       }
    }
    i++;
    while(i<a.size()){
      ans+=a[i]*fr[i];
      i++;
    }
    cout << ans << endl;
}

signed main(){
    fastio();
    #ifndef ONLINE_JUDGE   
      freopen("input.txt","r",stdin);
      freopen("output.txt","w",stdout);
    #endif 
    test
    solve();
    return 0;
}
### Prompt

In CPP, your task is to solve 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 <bits/stdc++.h>
using namespace std;
#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define ll  long long
#define int long long
#define ld  long double
#define endl "\n"
#define pb  push_back
#define fill(a,val) memset(a,val,sizeof(a))
#define ff  first
#define ss  second
#define test  ll t; cin>>t; while(t--)
#define loop(i,a,b)  for(ll i=a;i<b;i++)
#define loopr(i,a,b) for(ll i=a;i>=b;i--)
#define pii pair<ll,ll>
#define all(v) v.begin(),v.end()
istream& in(cin);
ostream& out(cout);
const ll mod     = 1000 * 1000 * 1000;
const ll inf     = 1ll*1000*1000*1000*1000*1000*1000 + 7;
const ll mod2    = 998244353;
const ll N       = 1000 * 1000 +100;
const ld pi      = 3.141592653589793;
ll power(ll x,ll y,ll p = LLONG_MAX ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}

void solve(){
    ll n,x;
    cin >> n >> x;
    vector<ll>a;
    vector<ll>fr;
    for(ll i=0;i<n;i++){
      ll y;
      cin >> y;
      a.pb(y);
      fr.pb(1);
    }
    ll ans=0,i;
    for(i=0;i<a.size();i++){
       if(a[i]%x==0){
         ll cnt=x*fr[i];
         a.pb(a[i]/x);
         fr.pb(cnt);
         ans+=fr[i]*a[i];
       }
       else{
         ans+=fr[i]*a[i];
         break;
       }
    }
    i++;
    while(i<a.size()){
      ans+=a[i]*fr[i];
      i++;
    }
    cout << ans << endl;
}

signed main(){
    fastio();
    #ifndef ONLINE_JUDGE   
      freopen("input.txt","r",stdin);
      freopen("output.txt","w",stdout);
    #endif 
    test
    solve();
    return 0;
}
```