#include <bits/stdc++.h>
using namespace std;
void fun() {}
long long int md = 1e9 + 7;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b, long long int md) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a) % md;
    }
    a = (1ll * a * a) % md;
    b >>= 1;
  }
  return res;
}
long long int poww(long long int a, long long int b) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a);
    }
    a = (1ll * a * a);
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
}
void disp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << "\n";
}
void dispv(vector<long long int> &v) {
  for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
long long int divide(long long int a, long long int b, long long int md) {
  long long int rr = a * (poww(b, md - 2, md));
  rr %= md;
  return rr;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fun();
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int a;
    map<long long int, long long int> mapp;
    for (long long int i = 1; i <= n; i++) {
      cin >> a;
      long long int md = a % k;
      long long int inc = (k - md) % k;
      mapp[inc]++;
    }
    long long int maxi = -1;
    for (auto i : mapp) {
      if (i.first == 0) continue;
      for (long long int j = 1; j < i.second; j++) mapp[i.first + j * k]++;
      mapp[i.first] = 1;
      maxi = max(maxi, i.first);
    }
    cout << maxi + 1 << "\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;
void fun() {}
long long int md = 1e9 + 7;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b, long long int md) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a) % md;
    }
    a = (1ll * a * a) % md;
    b >>= 1;
  }
  return res;
}
long long int poww(long long int a, long long int b) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a);
    }
    a = (1ll * a * a);
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
}
void disp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << "\n";
}
void dispv(vector<long long int> &v) {
  for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
long long int divide(long long int a, long long int b, long long int md) {
  long long int rr = a * (poww(b, md - 2, md));
  rr %= md;
  return rr;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fun();
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int a;
    map<long long int, long long int> mapp;
    for (long long int i = 1; i <= n; i++) {
      cin >> a;
      long long int md = a % k;
      long long int inc = (k - md) % k;
      mapp[inc]++;
    }
    long long int maxi = -1;
    for (auto i : mapp) {
      if (i.first == 0) continue;
      for (long long int j = 1; j < i.second; j++) mapp[i.first + j * k]++;
      mapp[i.first] = 1;
      maxi = max(maxi, i.first);
    }
    cout << maxi + 1 << "\n";
  }
  return 0;
}

```