#include <bits/stdc++.h>
using namespace std;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
int gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void SieveOfEratosthenes(int n, bool prime[], bool primesquare[], int a[]) {
  for (int i = 2; i <= n; i++) prime[i] = true;
  for (int i = 0; i <= (n * n + 1); i++) primesquare[i] = false;
  prime[1] = false;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  int j = 0;
  for (int p = 2; p <= n; p++) {
    if (prime[p]) {
      a[j] = p;
      primesquare[p * p] = true;
      j++;
    }
  }
}
int countDivisors(int n) {
  if (n == 1) return 1;
  bool prime[n + 1], primesquare[n * n + 1];
  int a[n];
  SieveOfEratosthenes(n, prime, primesquare, a);
  int ans = 1;
  for (int i = 0;; i++) {
    if (a[i] * a[i] * a[i] > n) break;
    int cnt = 1;
    while (n % a[i] == 0) {
      n = n / a[i];
      cnt = cnt + 1;
    }
    ans = ans * cnt;
  }
  if (prime[n])
    ans = ans * 2;
  else if (primesquare[n])
    ans = ans * 3;
  else if (n != 1)
    ans = ans * 4;
  return ans;
}
void div2() {
  long long t, n, x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long a[n];
    map<long long, long long> make_pair;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      make_pair[a[i] % k]++;
    }
    long long mx = -1;
    for (auto x : make_pair) {
      if (x.first != 0) {
        mx = std::max(mx, ((k - x.first) + (x.second - 1) * k));
      }
    }
    cout << mx + 1 << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
int gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void SieveOfEratosthenes(int n, bool prime[], bool primesquare[], int a[]) {
  for (int i = 2; i <= n; i++) prime[i] = true;
  for (int i = 0; i <= (n * n + 1); i++) primesquare[i] = false;
  prime[1] = false;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  int j = 0;
  for (int p = 2; p <= n; p++) {
    if (prime[p]) {
      a[j] = p;
      primesquare[p * p] = true;
      j++;
    }
  }
}
int countDivisors(int n) {
  if (n == 1) return 1;
  bool prime[n + 1], primesquare[n * n + 1];
  int a[n];
  SieveOfEratosthenes(n, prime, primesquare, a);
  int ans = 1;
  for (int i = 0;; i++) {
    if (a[i] * a[i] * a[i] > n) break;
    int cnt = 1;
    while (n % a[i] == 0) {
      n = n / a[i];
      cnt = cnt + 1;
    }
    ans = ans * cnt;
  }
  if (prime[n])
    ans = ans * 2;
  else if (primesquare[n])
    ans = ans * 3;
  else if (n != 1)
    ans = ans * 4;
  return ans;
}
void div2() {
  long long t, n, x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long a[n];
    map<long long, long long> make_pair;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      make_pair[a[i] % k]++;
    }
    long long mx = -1;
    for (auto x : make_pair) {
      if (x.first != 0) {
        mx = std::max(mx, ((k - x.first) + (x.second - 1) * k));
      }
    }
    cout << mx + 1 << "\n";
  }
}

```