#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long mod1 = 998244353;
unsigned long long power(unsigned long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else {
    if (y % 2 == 0)
      return power(x * x, y / 2);
    else
      return x * power(x * x, (y - 1) / 2);
  }
}
long long mod_power(long long x, long long y, long long m) {
  long long r = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) r = (r * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return r;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
vector<long long> extended_Euclid(long long a, long long b) {
  vector<long long> v(3);
  if (b == 0) {
    v[0] = a;
    v[1] = 1;
    v[2] = 0;
    return v;
  } else {
    vector<long long> v1 = extended_Euclid(b, a % b);
    v[0] = v1[0];
    v[1] = v1[2];
    v[2] = v1[1] - (a / b) * v1[2];
    return v;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    map<long long, long long> M;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      v[i] = v[i] % k;
      if (v[i] != 0) M[v[i]] += 1;
    }
    if (M.size() > 0) {
      long long x = 0;
      long long idx = 0;
      auto ip = M.begin();
      for (ip; ip != M.end(); ip++) {
        if (ip->second > idx) {
          idx = ip->second;
          x = ip->first;
        }
      }
      long long j = k - x;
      long long ans = (idx - 1) * k + j;
      cout << ans + 1 << endl;
    } else
      cout << 0 << endl;
  }
}

### 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;
const long long mod = 1e9 + 7;
const long long mod1 = 998244353;
unsigned long long power(unsigned long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else {
    if (y % 2 == 0)
      return power(x * x, y / 2);
    else
      return x * power(x * x, (y - 1) / 2);
  }
}
long long mod_power(long long x, long long y, long long m) {
  long long r = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) r = (r * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return r;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
vector<long long> extended_Euclid(long long a, long long b) {
  vector<long long> v(3);
  if (b == 0) {
    v[0] = a;
    v[1] = 1;
    v[2] = 0;
    return v;
  } else {
    vector<long long> v1 = extended_Euclid(b, a % b);
    v[0] = v1[0];
    v[1] = v1[2];
    v[2] = v1[1] - (a / b) * v1[2];
    return v;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    map<long long, long long> M;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      v[i] = v[i] % k;
      if (v[i] != 0) M[v[i]] += 1;
    }
    if (M.size() > 0) {
      long long x = 0;
      long long idx = 0;
      auto ip = M.begin();
      for (ip; ip != M.end(); ip++) {
        if (ip->second > idx) {
          idx = ip->second;
          x = ip->first;
        }
      }
      long long j = k - x;
      long long ans = (idx - 1) * k + j;
      cout << ans + 1 << endl;
    } else
      cout << 0 << endl;
  }
}

```