You are given two arrays of integers a_1, a_2, …, a_n and b_1, b_2, …, b_m.

You need to insert all elements of b into a in an arbitrary way. As a result you will get an array c_1, c_2, …, c_{n+m} of size n + m.

Note that you are not allowed to change the order of elements in a, while you can insert elements of b at arbitrary positions. They can be inserted at the beginning, between any elements of a, or at the end. Moreover, elements of b can appear in the resulting array in any order.

What is the minimum possible number of inversions in the resulting array c? Recall that an inversion is a pair of indices (i, j) such that i < j and c_i > c_j.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.

The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^6).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).

The third line of each test case contains m integers b_1, b_2, …, b_m (1 ≤ b_i ≤ 10^9).

It is guaranteed that the sum of n for all tests cases in one input doesn't exceed 10^6. The sum of m for all tests cases doesn't exceed 10^6 as well.

Output

For each test case, print one integer — the minimum possible number of inversions in the resulting array c.

Example

Input


3
3 4
1 2 3
4 3 2 1
3 3
3 2 1
1 2 3
5 4
1 3 5 3 1
4 3 6 1


Output


0
4
6

Note

Below is given the solution to get the optimal answer for each of the example test cases (elements of a are underscored).

  * In the first test case, c = [\underline{1}, 1, \underline{2}, 2, \underline{3}, 3, 4]. 
  * In the second test case, c = [1, 2, \underline{3}, \underline{2}, \underline{1}, 3]. 
  * In the third test case, c = [\underline{1}, 1, 3, \underline{3}, \underline{5}, \underline{3}, \underline{1}, 4, 6]. 
#include <bits/stdc++.h>
using namespace std;
void per() { cerr << endl; }
template <typename Head, typename... Tail>
void per(Head H, Tail... T) {
  cerr << H << ' ';
  per(T...);
}
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& a) {
  return out << "(" << a.first << ", " << a.second << ")";
}
template <class U, class V>
istream& operator>>(istream& in, pair<U, V>& a) {
  return in >> a.first >> a.second;
}
template <typename W,
          typename T = typename enable_if<!is_same<W, string>::value,
                                          typename W::value_type>::type>
ostream& operator<<(ostream& out, const W& v) {
  out << "{ ";
  for (const auto& first : v) out << first << ", ";
  return out << '}';
}
template <class T>
void readArr(T from, T to) {
  for (auto i = from; i != to; ++i) cin >> *i;
}
mt19937 mrand(1337);
unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
unsigned long long myRand64() {
  return ((unsigned long long)myRand32() << 32) ^ myRand32();
}
const int mod = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void dec(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return a * (long long)b % mod; }
int bp(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mult(res, a);
    a = mult(a, a);
    b >>= 1;
  }
  return res;
}
struct Vertex {
  int sum;
  int min_prefix_sum;
};
vector<Vertex> rmq;
void build(int i, int l, int r) {
  if (r - l == 1) {
    rmq[i].sum = 1;
    rmq[i].min_prefix_sum = 0;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * i + 1, l, mid);
  build(2 * i + 2, mid, r);
  rmq[i].sum = rmq[2 * i + 1].sum + rmq[2 * i + 2].sum;
  rmq[i].min_prefix_sum =
      min(rmq[2 * i + 1].min_prefix_sum,
          rmq[2 * i + 1].sum + rmq[2 * i + 2].min_prefix_sum);
}
void upd(int i, int l, int r, int index) {
  if (r - l == 1) {
    rmq[i].sum = -1;
    rmq[i].min_prefix_sum = -1;
    return;
  }
  int mid = (l + r) / 2;
  if (index < mid)
    upd(2 * i + 1, l, mid, index);
  else
    upd(2 * i + 2, mid, r, index);
  rmq[i].sum = rmq[2 * i + 1].sum + rmq[2 * i + 2].sum;
  rmq[i].min_prefix_sum =
      min(rmq[2 * i + 1].min_prefix_sum,
          rmq[2 * i + 1].sum + rmq[2 * i + 2].min_prefix_sum);
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > farr;
  vector<int> sarr(m);
  for (int i = 0; i < n; ++i) {
    int first;
    cin >> first;
    farr.push_back(make_pair(first, i));
  }
  for (auto& first : sarr) cin >> first;
  sort(farr.begin(), farr.end());
  sort(sarr.begin(), sarr.end());
  rmq.assign(4 * n, {});
  build(0, 0, n);
  int cur = 0;
  long long ans = 0;
  int u = 0;
  for (auto first : sarr) {
    while (u < farr.size() && farr[u].first <= first) {
      cur++;
      upd(0, 0, n, farr[u++].second);
    }
    ans += cur + rmq[0].min_prefix_sum;
  }
  cout << 2 * ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
}
