#include <bits/stdc++.h>
using namespace std;
template <typename H>
bool chmin(H& v1, const H v2) {
  if (v1 > v2) {
    v1 = v2;
    return true;
  }
  return false;
}
template <typename H>
bool chmax(H& v1, const H v2) {
  if (v1 < v2) {
    v1 = v2;
    return true;
  }
  return false;
}
template <typename H>
void read(H& head) {
  cin >> head;
}
template <typename H, typename... T>
void read(H& head, T&... tail) {
  cin >> head;
  read(tail...);
}
template <typename H>
void write(H head) {
  cout << head << '\n';
}
template <typename H, typename... T>
void write(H head, T... tail) {
  cout << head << " ";
  write(tail...);
}
template <typename... T>
void die(T... tok) {
  write(tok...);
  exit(0);
}
void solve() {
  int n, k;
  read(n, k);
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) {
    read(a[i]);
    a[i] = (k - a[i] % k) % k;
  }
  multiset<int> m(begin(a), end(a));
  while (m.find(0) != m.end()) m.erase(m.find(0));
  long long cur = 0;
  while (m.size()) {
    auto it = m.upper_bound(cur % k);
    if (it == m.end()) {
      it = m.begin();
      cur += k - cur % k;
    }
    cur += (*it - cur % k);
    m.erase(it);
  }
  if (cur) ++cur;
  write(cur);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  read(t);
  for (int i = 0; i < (t); ++i) solve();
}

### Prompt

Please formulate 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;
template <typename H>
bool chmin(H& v1, const H v2) {
  if (v1 > v2) {
    v1 = v2;
    return true;
  }
  return false;
}
template <typename H>
bool chmax(H& v1, const H v2) {
  if (v1 < v2) {
    v1 = v2;
    return true;
  }
  return false;
}
template <typename H>
void read(H& head) {
  cin >> head;
}
template <typename H, typename... T>
void read(H& head, T&... tail) {
  cin >> head;
  read(tail...);
}
template <typename H>
void write(H head) {
  cout << head << '\n';
}
template <typename H, typename... T>
void write(H head, T... tail) {
  cout << head << " ";
  write(tail...);
}
template <typename... T>
void die(T... tok) {
  write(tok...);
  exit(0);
}
void solve() {
  int n, k;
  read(n, k);
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) {
    read(a[i]);
    a[i] = (k - a[i] % k) % k;
  }
  multiset<int> m(begin(a), end(a));
  while (m.find(0) != m.end()) m.erase(m.find(0));
  long long cur = 0;
  while (m.size()) {
    auto it = m.upper_bound(cur % k);
    if (it == m.end()) {
      it = m.begin();
      cur += k - cur % k;
    }
    cur += (*it - cur % k);
    m.erase(it);
  }
  if (cur) ++cur;
  write(cur);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  read(t);
  for (int i = 0; i < (t); ++i) solve();
}

```