#include <bits/stdc++.h>
using namespace std;
long long fastexp(long long b, long long e) {
  long long ans = 1;
  long long lpow = b;
  while (e) {
    if (e & 1) ans *= lpow;
    lpow *= lpow;
    e = (e >> 1);
  }
  return ans;
}
bool isprime(long long x) {
  for (long long i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
void solve() {
  long long ele, n, k, lv, ans = 0;
  long long maxr = -1, mocc = 0;
  bool need = false;
  cin >> n >> k;
  map<long long, long long> um;
  um.clear();
  for (lv = 1; lv <= n; lv++) {
    cin >> ele;
    if (ele % k) {
      need = true;
      ++um[k - (ele % k)];
      mocc = max(mocc, um[k - (ele % k)]);
    }
  }
  if (need == false) {
    cout << 0;
    return;
  }
  maxr = -1;
  for (auto it = um.rbegin(); it != um.rend(); it++) {
    if (it->second == mocc) {
      maxr = it->first;
      break;
    }
  }
  ans = (mocc - 1) * k + (maxr + 1);
  cout << ans;
}
int main() {
  long long tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> tc;
  while (tc--) {
    solve();
    cout << "\n";
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

You are given an array a consisting of n positive integers.

Initially, you have an integer x = 0. During one move, you can do one of the following two operations:

  1. Choose exactly one i from 1 to n and increase a_i by x (a_i := a_i + x), then increase x by 1 (x := x + 1). 
  2. Just increase x by 1 (x := x + 1). 



The first operation can be applied no more than once to each i from 1 to n.

Your task is to find the minimum number of moves required to obtain such an array that each its element is divisible by k (the value k is given).

You have to answer t independent test cases. 

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of a and the required divisior. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of moves required to obtain such an array that each its element is divisible by k.

Example

Input


5
4 3
1 2 1 3
10 6
8 7 1 8 3 7 5 10 8 9
5 10
20 100 50 20 100500
10 25
24 24 24 24 24 24 24 24 24 24
8 8
1 2 3 4 5 6 7 8


Output


6
18
0
227
8

Note

Consider the first test case of the example:

  1. x=0, a = [1, 2, 1, 3]. Just increase x; 
  2. x=1, a = [1, 2, 1, 3]. Add x to the second element and increase x; 
  3. x=2, a = [1, 3, 1, 3]. Add x to the third element and increase x; 
  4. x=3, a = [1, 3, 3, 3]. Add x to the fourth element and increase x; 
  5. x=4, a = [1, 3, 3, 6]. Just increase x; 
  6. x=5, a = [1, 3, 3, 6]. Add x to the first element and increase x; 
  7. x=6, a = [6, 3, 3, 6]. We obtained the required array. 



Note that you can't add x to the same element more than once.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long fastexp(long long b, long long e) {
  long long ans = 1;
  long long lpow = b;
  while (e) {
    if (e & 1) ans *= lpow;
    lpow *= lpow;
    e = (e >> 1);
  }
  return ans;
}
bool isprime(long long x) {
  for (long long i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
void solve() {
  long long ele, n, k, lv, ans = 0;
  long long maxr = -1, mocc = 0;
  bool need = false;
  cin >> n >> k;
  map<long long, long long> um;
  um.clear();
  for (lv = 1; lv <= n; lv++) {
    cin >> ele;
    if (ele % k) {
      need = true;
      ++um[k - (ele % k)];
      mocc = max(mocc, um[k - (ele % k)]);
    }
  }
  if (need == false) {
    cout << 0;
    return;
  }
  maxr = -1;
  for (auto it = um.rbegin(); it != um.rend(); it++) {
    if (it->second == mocc) {
      maxr = it->first;
      break;
    }
  }
  ans = (mocc - 1) * k + (maxr + 1);
  cout << ans;
}
int main() {
  long long tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> tc;
  while (tc--) {
    solve();
    cout << "\n";
  }
}

```