#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void normal(long long int &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long int modMul(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long int modSub(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long int modInverse(long long int a) {
  return modPow(a, 1000000007 - 2);
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modInverse(b));
}
inline bool isInside(pair<int, int> p, long long int n, long long int m) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < m);
}
inline bool isInside(pair<int, int> p, long long int n) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < n);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isFib(long long int x) {
  return isSquare(5 * x * x + 4) || isSquare(5 * x * x - 4);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << endl; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
struct func {
  bool operator()(pair<int, int> const &a, pair<int, int> const &b) {
    if (a.first == b.first) return (a.second < b.second);
    return (a.first < b.first);
  }
};
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const int mx = (int)1e5 + 9;
long long int n, m, a, b, t, i, j, d, cs = 0, counT = 0, k, ans = 0, l = 0,
                                      sum1 = 0, sum = 0, Max, Min, num;
int main() {
  ios_base::sync_with_stdio(false);
  t = 1;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long int arr[n + 9];
    map<long long int, long long int> mp;
    Max = 0;
    for (int i = 0; i < (n); i++) {
      cin >> arr[i];
      arr[i] = arr[i] % k;
      if (arr[i]) {
        mp[k - arr[i]]++;
      }
    }
    for (auto it : mp) {
      d = (it.second - 1LL) * k + it.first;
      Max = max(Max, d);
    }
    if (Max) Max++;
    cout << Max << '\n';
  }
}

### Prompt

Please create a solution in CPP 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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void normal(long long int &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long int modMul(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long int modSub(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long int modInverse(long long int a) {
  return modPow(a, 1000000007 - 2);
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modInverse(b));
}
inline bool isInside(pair<int, int> p, long long int n, long long int m) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < m);
}
inline bool isInside(pair<int, int> p, long long int n) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < n);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isFib(long long int x) {
  return isSquare(5 * x * x + 4) || isSquare(5 * x * x - 4);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << endl; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
struct func {
  bool operator()(pair<int, int> const &a, pair<int, int> const &b) {
    if (a.first == b.first) return (a.second < b.second);
    return (a.first < b.first);
  }
};
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const int mx = (int)1e5 + 9;
long long int n, m, a, b, t, i, j, d, cs = 0, counT = 0, k, ans = 0, l = 0,
                                      sum1 = 0, sum = 0, Max, Min, num;
int main() {
  ios_base::sync_with_stdio(false);
  t = 1;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long int arr[n + 9];
    map<long long int, long long int> mp;
    Max = 0;
    for (int i = 0; i < (n); i++) {
      cin >> arr[i];
      arr[i] = arr[i] % k;
      if (arr[i]) {
        mp[k - arr[i]]++;
      }
    }
    for (auto it : mp) {
      d = (it.second - 1LL) * k + it.first;
      Max = max(Max, d);
    }
    if (Max) Max++;
    cout << Max << '\n';
  }
}

```