#include <bits/stdc++.h>
using namespace std;
void mer(int *a, int *l, int ln, int *r, int rn) {
  int i = 0, j = 0, k = 0;
  while (i < ln && j < rn) {
    if (l[i] < r[j])
      a[k++] = l[i++];
    else
      a[k++] = r[j++];
  }
  while (i < ln) a[k++] = l[i++];
  while (j < rn) a[k++] = r[j++];
}
void mers(int *a, int n) {
  if (n < 2) return;
  int i, mid = n / 2;
  int l[mid], r[n - mid];
  for (i = 0; i < mid; i++) l[i] = a[i];
  for (i = mid; i < n; i++) r[i - mid] = a[i];
  mers(l, mid);
  mers(r, n - mid);
  mer(a, l, mid, r, n - mid);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int x = 0, m, n, k, pk = 0, px = 0, z = 0, l = 0;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> m;
      a[i] = m % k;
    }
    m = 0;
    mers(a, n);
    for (int i = n - 1; i >= 0; i--) {
      if (z == 0) {
        if (a[i] != 0) {
          if ((a[i] + x) % k == 0)
            x += 1;
          else
            x = x + (k - (x + a[i]) % k) + 1;
        }
      }
      while (a[i] == a[i - 1] && a[i] != 0 && i >= 0) {
        z = 1;
        i--;
        l++;
        if (l >= pk) {
          if ((a[i] + x) % k == 0)
            x += 1;
          else
            x = x + (k - (x + a[i]) % k) + 1;
        }
      }
      if (l > pk) {
        pk = l;
        l = 0;
      } else
        l = 0;
      px = x;
    }
    cout << x << endl;
  }
}

### 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 mer(int *a, int *l, int ln, int *r, int rn) {
  int i = 0, j = 0, k = 0;
  while (i < ln && j < rn) {
    if (l[i] < r[j])
      a[k++] = l[i++];
    else
      a[k++] = r[j++];
  }
  while (i < ln) a[k++] = l[i++];
  while (j < rn) a[k++] = r[j++];
}
void mers(int *a, int n) {
  if (n < 2) return;
  int i, mid = n / 2;
  int l[mid], r[n - mid];
  for (i = 0; i < mid; i++) l[i] = a[i];
  for (i = mid; i < n; i++) r[i - mid] = a[i];
  mers(l, mid);
  mers(r, n - mid);
  mer(a, l, mid, r, n - mid);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int x = 0, m, n, k, pk = 0, px = 0, z = 0, l = 0;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> m;
      a[i] = m % k;
    }
    m = 0;
    mers(a, n);
    for (int i = n - 1; i >= 0; i--) {
      if (z == 0) {
        if (a[i] != 0) {
          if ((a[i] + x) % k == 0)
            x += 1;
          else
            x = x + (k - (x + a[i]) % k) + 1;
        }
      }
      while (a[i] == a[i - 1] && a[i] != 0 && i >= 0) {
        z = 1;
        i--;
        l++;
        if (l >= pk) {
          if ((a[i] + x) % k == 0)
            x += 1;
          else
            x = x + (k - (x + a[i]) % k) + 1;
        }
      }
      if (l > pk) {
        pk = l;
        l = 0;
      } else
        l = 0;
      px = x;
    }
    cout << x << endl;
  }
}

```