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;
int t, n, m;
pair<int, int> a[1000005], b[1000005];
int bit[2 * 1000005 + 1];
int sum(int i) {
  int s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & -i;
  }
  return s;
}
void add(int i, int x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
struct segtree {
  int ans[4 * 1000005], sum[4 * 1000005], rsum[4 * 1000005];
  void pushup(int k) {
    ans[k] = min(ans[k * 2] + rsum[k * 2 + 1], ans[k * 2 + 1] + sum[k * 2]);
    sum[k] = sum[k * 2] + sum[k * 2 + 1];
    rsum[k] = rsum[k * 2] + rsum[k * 2 + 1];
  }
  void build(int k, int l, int r) {
    if (l == r) {
      sum[k] = 1;
      rsum[k] = 0;
      ans[k] = 0;
      return;
    }
    int mid = (l + r) / 2;
    build(k * 2, l, mid);
    build(k * 2 + 1, mid + 1, r);
    pushup(k);
  }
  void update(int k, int l, int r, int p, int type) {
    if (l == r) {
      if (type == 0) {
        sum[k] = 0;
        rsum[k] = 0;
      } else {
        sum[k] = 0;
        rsum[k] = 1;
      }
      return;
    }
    int mid = (l + r) / 2;
    if (p <= mid)
      update(k * 2, l, mid, p, type);
    else
      update(k * 2 + 1, mid + 1, r, p, type);
    pushup(k);
  }
} seg;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i].first);
      a[i].second = i;
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d", &b[i].first);
      b[i].second = i;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) bit[i] = 0;
    sort(a + 1, a + n + 1);
    sort(b + 1, b + m + 1);
    int l = 1;
    seg.build(1, 1, n);
    vector<int> equal;
    for (int i = 1; i <= m; i++) {
      if (i == 1 || b[i].first != b[i - 1].first) {
        for (auto x : equal) seg.update(1, 1, n, x, 1);
        equal.clear();
      } else {
        ans += seg.ans[1];
        continue;
      }
      while (l <= n && a[l].first <= b[i].first) {
        if (a[l].first == b[i].first) {
          equal.push_back(a[l].second);
          seg.update(1, 1, n, a[l].second, 0);
        } else
          seg.update(1, 1, n, a[l].second, 1);
        l++;
      }
      ans += seg.ans[1];
    }
    int now = 1;
    while (now <= n) {
      int cur = now;
      while (cur + 1 <= n && a[cur].first == a[cur + 1].first) cur++;
      for (int i = now; i <= cur; i++) ans += sum(n + 1 - a[i].second);
      for (int i = now; i <= cur; i++) add(n + 1 - a[i].second, 1);
      now = cur + 1;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
