#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <string>
#include <vector>

typedef long long ll;
typedef unsigned long long ull;
typedef long double LD;
typedef std::pair<int, ll> pii;
const int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const ll INF64 = (ll)1e18;
const double INFD = 1e30;
const double EPS = 1e-9;
const double PI = std::acos(-1);
const int MOD = 1e9 + 7;
template <typename T>
inline T read() {
    T X = 0, w = 0;
    char ch = 0;
    while (!isdigit(ch)) {
        w |= ch == '-';
        ch = getchar();
    }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
const int MAXN = 100005 * 64;
const int MAXV = (1 << 17) + 5;
int n, m, k;
int dr[4] = {0, 1, 0, -1};
int dc[4] = {-1, 0, 1, 0};
int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
int CASE = 1;
pii A[MAXN];
char S[MAXN];
void solve() {
    scanf("%d%d", &n, &m);
    int end = n;
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
        A[i].first = read<int>();
        A[i].second = 1;
        sum += A[i].first;
    }
    for (int i = 1; i <= end; i++) {
        if (A[i].first % m == 0) {
            A[++end] = {A[i].first / m, (ll)A[i].second * m};
            sum += (ll)A[i].first * A[i].second;
        } else {
            break;
        }
    }
    printf("%lld\n", sum);
}
int main() {
#ifdef LOCALLL
    freopen("in", "r", stdin);
    freopen("out", "w", stdout);
#endif
    int T = read<int>();
    while (T--) {
        solve();
    }
    return 0;
}
### 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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <string>
#include <vector>

typedef long long ll;
typedef unsigned long long ull;
typedef long double LD;
typedef std::pair<int, ll> pii;
const int INF = 2147483647;
const int INF2 = 0x3f3f3f3f;
const ll INF64 = (ll)1e18;
const double INFD = 1e30;
const double EPS = 1e-9;
const double PI = std::acos(-1);
const int MOD = 1e9 + 7;
template <typename T>
inline T read() {
    T X = 0, w = 0;
    char ch = 0;
    while (!isdigit(ch)) {
        w |= ch == '-';
        ch = getchar();
    }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
const int MAXN = 100005 * 64;
const int MAXV = (1 << 17) + 5;
int n, m, k;
int dr[4] = {0, 1, 0, -1};
int dc[4] = {-1, 0, 1, 0};
int dr2[8] = {1, 1, 1, -1, -1, -1, 0, 0};
int dc2[8] = {1, 0, -1, 1, 0, -1, 1, -1};
int CASE = 1;
pii A[MAXN];
char S[MAXN];
void solve() {
    scanf("%d%d", &n, &m);
    int end = n;
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
        A[i].first = read<int>();
        A[i].second = 1;
        sum += A[i].first;
    }
    for (int i = 1; i <= end; i++) {
        if (A[i].first % m == 0) {
            A[++end] = {A[i].first / m, (ll)A[i].second * m};
            sum += (ll)A[i].first * A[i].second;
        } else {
            break;
        }
    }
    printf("%lld\n", sum);
}
int main() {
#ifdef LOCALLL
    freopen("in", "r", stdin);
    freopen("out", "w", stdout);
#endif
    int T = read<int>();
    while (T--) {
        solve();
    }
    return 0;
}
```