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;
const int maxn = 2100020;
const int maxm = 10000020;
const int MOd = 998244353;
int a, b, ar[maxn], bs[maxn], sz[maxn * 2];
int n, segment[maxn * 2], lazy[maxn * 2];
map<int, int> mp;
vector<int> w[maxn];
void push(int k) {
  if (lazy[k]) {
    if (k < n) {
      lazy[k + k] += lazy[k];
      lazy[k + k + 1] += lazy[k];
    }
    segment[k] += lazy[k];
    lazy[k] = 0;
  }
}
void up(int k, int b, int s, int x, int y, int m) {
  push(k);
  if (b > y || x > s) return;
  if (x <= b && s <= y) {
    lazy[k] += m;
    push(k);
    return;
  }
  up(k + k, b, (b + s) / 2, x, y, m);
  up(k + k + 1, (b + s) / 2 + 1, s, x, y, m);
  segment[k] = min(segment[k + k], segment[k + k + 1]);
}
int fw[maxn], cnt;
int find(int x) {
  int ret = 0;
  for (int i = x; i <= cnt; i += i & (-i)) ret += fw[i];
  return ret;
}
void fup(int x) {
  for (int i = x; i > 0; i -= i & (-i)) fw[i] += 1;
}
void solve() {
  scanf("%d %d", &a, &b);
  mp.clear();
  for (int i = 1; i <= a; i++) scanf("%d", &ar[i]), mp[ar[i]] = 1;
  for (int i = 1; i <= b; i++) scanf("%d", &bs[i]), mp[bs[i]] = 1;
  cnt = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
    it->second = ++cnt;
  n = 1;
  while (n < a + 1) n <<= 1;
  for (int i = 1; i <= n + n; i++) segment[i] = lazy[i] = sz[i] = 0;
  for (int i = 0; i <= cnt; i++) fw[i] = 0;
  for (int i = 1; i <= a; i++)
    w[mp[ar[i]]].push_back(i), up(1, 1, n, i + 1, n, 1);
  for (int i = 1; i <= b; i++) sz[mp[bs[i]]] += 1;
  long long ans = 0;
  for (int i = 1; i <= a; i++) {
    ans += find(mp[ar[i]] + 1);
    fup(mp[ar[i]]);
  }
  for (int i = 1; i <= cnt; i++) {
    for (int j = 0; j < w[i].size(); j++) up(1, 1, n, w[i][j] + 1, n, -1);
    push(1);
    ans += (long long)segment[1] * sz[i];
    for (int j = 0; j < w[i].size(); j++) up(1, 1, n, 1, w[i][j], 1);
  }
  cout << ans << endl;
  for (int i = 1; i <= cnt; i++) w[i].clear();
}
int main() {
  int n = 1;
  scanf("%d", &n);
  while (n--) solve();
  return 0;
}
