#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 1e9 + 7;
const long long sz = 2e5 + 9;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void bring_me_thanos() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long &x : v) {
    cin >> x;
    x %= k;
    x = (k - x + k) % k;
  }
  long long c = 0, f = 1;
  map<long long, long long> ct;
  for (auto x : v) {
    if (x == 0) {
      continue;
    }
    f = 0;
    c = max(ct[x] * k + x, c);
    ct[x]++;
  }
  if (f) {
    cout << 0 << '\n';
    return;
  }
  cout << c + 1 << '\n';
}
void solve() {
  long long tc;
  cin >> tc;
  while (tc--) bring_me_thanos();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t beg = clock();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec, Copyright %c 2019 PyThor. \n",
          double(end - beg) / CLOCKS_PER_SEC, 184);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 1e9 + 7;
const long long sz = 2e5 + 9;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void bring_me_thanos() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long &x : v) {
    cin >> x;
    x %= k;
    x = (k - x + k) % k;
  }
  long long c = 0, f = 1;
  map<long long, long long> ct;
  for (auto x : v) {
    if (x == 0) {
      continue;
    }
    f = 0;
    c = max(ct[x] * k + x, c);
    ct[x]++;
  }
  if (f) {
    cout << 0 << '\n';
    return;
  }
  cout << c + 1 << '\n';
}
void solve() {
  long long tc;
  cin >> tc;
  while (tc--) bring_me_thanos();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t beg = clock();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec, Copyright %c 2019 PyThor. \n",
          double(end - beg) / CLOCKS_PER_SEC, 184);
  return 0;
}

```