#include <bits/stdc++.h>
using namespace std;
bool Prime(long long int n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    long long int n, k;
    cin >> n >> k;
    ;
    vector<long long int> ar;
    for (long long int i = 0; i < n; i++) {
      long long int a;
      cin >> a;
      ;
      ar.push_back(a);
    }
    map<long long int, long long int> make_pair;
    long long int p = 0;
    for (long long int i = 0; i <= ar.size() - 1; i++) {
      if (ar[i] % k) {
        long long int d = k - (ar[i] % k);
        if (make_pair[d] == 0) {
          p = max(p, d);
          make_pair[d]++;
        } else {
          long long int s = make_pair[d];
          while (make_pair[d + (k * s)] != 0) s++;
          p = max(p, d + (s * k));
          make_pair[d] = s + 1;
          make_pair[d + (k * s)]++;
        }
      }
    }
    if (p)
      cout << p + 1 << "\n";
    else
      cout << p << "\n";
    ;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
bool Prime(long long int n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) {
    long long int n, k;
    cin >> n >> k;
    ;
    vector<long long int> ar;
    for (long long int i = 0; i < n; i++) {
      long long int a;
      cin >> a;
      ;
      ar.push_back(a);
    }
    map<long long int, long long int> make_pair;
    long long int p = 0;
    for (long long int i = 0; i <= ar.size() - 1; i++) {
      if (ar[i] % k) {
        long long int d = k - (ar[i] % k);
        if (make_pair[d] == 0) {
          p = max(p, d);
          make_pair[d]++;
        } else {
          long long int s = make_pair[d];
          while (make_pair[d + (k * s)] != 0) s++;
          p = max(p, d + (s * k));
          make_pair[d] = s + 1;
          make_pair[d + (k * s)]++;
        }
      }
    }
    if (p)
      cout << p + 1 << "\n";
    else
      cout << p << "\n";
    ;
  }
  return 0;
}

```