#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long int ull;
typedef long long ll;
typedef double db;

const ll mod = 1e9+7;
const db eps = 1e-9;

#define MAX 1000000000
#define MIN 100000
#define mp make_pair
#define pb push_back
#define endl "\n"
#define deb(x) cout << #x << " " << x << endl


int main()
{
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
        
    ll t;
    cin>>t;
    while(t--){
        ll n,x;
        ll su=0;
        cin>>n>>x;
        std::vector<ll> v(n);
        for(ll i=0 ; i<n ; i++){
            cin>>v[i];
            su = su + ull(v[i]);
        }
        int mi = 100000;
        std::vector<int> dp(n, 0);
        for(ll i=0  ; i<n  ;i++){
            ll tmp = v[i];
            if(tmp<x)break;
            int cnt=0;
            while((tmp%x)==0ll){
                cnt+=1;
                tmp/=x;
            }
            // if(cnt==0){
            //     break;
            // }
            // else{
            //     mi = min(mi, cnt);
            //     su = su + ull(mi*v[i]);
            //     // su+=(mi*v[i]);
            // }
            mi = min(cnt, mi);
            dp[i]=cnt;
        }
        bool res = false;
        for(ll i=0 ; i<n ; i++){
            if(dp[i]==0){
                break;
            }
            if(dp[i]==mi){
                res = true;
            }
            if(res){
                su = su + (mi*v[i]);
            }
            if(!res){
                su = su + ((mi+1)*v[i]);
            }
            // ll tmp = v[i];
            // if(tmp<x)break;
            // int cnt=0;
            // while((tmp%x)==0ll){
            //     cnt+=1;
            //     tmp/=x;
            // }
            // if(cnt==0){
            //     break;
            // }
            // else{
            //     mi = min(mi, cnt);
            //     su = su + ull(mi*v[i]);
                // su+=(mi*v[i]);
            // }
            // dp[i] = cnt;
        }
        // if(dp[0]==0){
        //     cout<<su<<endl;
        //     continue;
        // }
        // // ll index = -1;
        // for(ll i=0 ;i<n ; i++){
        //     if(dp[i]==0){
        //         // index = i;
        //         break;
        //     }
        //     else{
        //         mi = min(mi, dp[i]);
        //     }
        // }
        // bool res = true;
        // ll tmi;
        // for(ll i=0 ; i<n ; i++){
        //     if((v[i]%x) !=0){
        //         break;
        //     }
        //     if(dp[i]==mi){
        //         res = false;
        //     }
        //     if(!res){
        //         tmi = mi;
        //     }
        //     else{
        //         tmi = min(mi+1, dp[i]);
        //     }
        //     while(tmi>0){
        //         su+=v[i];
        //         --tmi;
        //     }
        // }
        cout<<su<<endl;
    }
    
    
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
#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long int ull;
typedef long long ll;
typedef double db;

const ll mod = 1e9+7;
const db eps = 1e-9;

#define MAX 1000000000
#define MIN 100000
#define mp make_pair
#define pb push_back
#define endl "\n"
#define deb(x) cout << #x << " " << x << endl


int main()
{
    ios_base :: sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
        
    ll t;
    cin>>t;
    while(t--){
        ll n,x;
        ll su=0;
        cin>>n>>x;
        std::vector<ll> v(n);
        for(ll i=0 ; i<n ; i++){
            cin>>v[i];
            su = su + ull(v[i]);
        }
        int mi = 100000;
        std::vector<int> dp(n, 0);
        for(ll i=0  ; i<n  ;i++){
            ll tmp = v[i];
            if(tmp<x)break;
            int cnt=0;
            while((tmp%x)==0ll){
                cnt+=1;
                tmp/=x;
            }
            // if(cnt==0){
            //     break;
            // }
            // else{
            //     mi = min(mi, cnt);
            //     su = su + ull(mi*v[i]);
            //     // su+=(mi*v[i]);
            // }
            mi = min(cnt, mi);
            dp[i]=cnt;
        }
        bool res = false;
        for(ll i=0 ; i<n ; i++){
            if(dp[i]==0){
                break;
            }
            if(dp[i]==mi){
                res = true;
            }
            if(res){
                su = su + (mi*v[i]);
            }
            if(!res){
                su = su + ((mi+1)*v[i]);
            }
            // ll tmp = v[i];
            // if(tmp<x)break;
            // int cnt=0;
            // while((tmp%x)==0ll){
            //     cnt+=1;
            //     tmp/=x;
            // }
            // if(cnt==0){
            //     break;
            // }
            // else{
            //     mi = min(mi, cnt);
            //     su = su + ull(mi*v[i]);
                // su+=(mi*v[i]);
            // }
            // dp[i] = cnt;
        }
        // if(dp[0]==0){
        //     cout<<su<<endl;
        //     continue;
        // }
        // // ll index = -1;
        // for(ll i=0 ;i<n ; i++){
        //     if(dp[i]==0){
        //         // index = i;
        //         break;
        //     }
        //     else{
        //         mi = min(mi, dp[i]);
        //     }
        // }
        // bool res = true;
        // ll tmi;
        // for(ll i=0 ; i<n ; i++){
        //     if((v[i]%x) !=0){
        //         break;
        //     }
        //     if(dp[i]==mi){
        //         res = false;
        //     }
        //     if(!res){
        //         tmi = mi;
        //     }
        //     else{
        //         tmi = min(mi+1, dp[i]);
        //     }
        //     while(tmi>0){
        //         su+=v[i];
        //         --tmi;
        //     }
        // }
        cout<<su<<endl;
    }
    
    
return 0;
}

```