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 M = 2e6 + 5;
int n, m, tmp2[M];
int a[M], b[M], c[M], pos[M];
void mymid(int ta[], int tb[], int tpos[], int l, int r, int L, int R) {
  if (L > R) return;
  int mid = (L + R) / 2;
  int now = 0, po, mmin;
  for (int i = l; i < r; i++)
    if (ta[i] < tb[mid]) now++;
  mmin = now, po = l;
  for (int i = l + 1; i <= r; i++) {
    if (ta[i - 1] < tb[mid]) now--;
    if (ta[i - 1] > tb[mid]) now++;
    if (now < mmin) mmin = now, po = i;
  }
  tpos[mid] = po;
  mymid(ta, tb, tpos, l, po, L, mid - 1);
  mymid(ta, tb, tpos, po, r, mid + 1, R);
}
void print(int tmp[], int len) {
  for (int i = 1; i <= len; i++) printf("%d ", tmp[i]);
  printf("\n");
}
long long mysort(int tmp[], int l, int r) {
  if (l >= r) return 0;
  long long mid = (l + r) / 2, res = 0;
  res += mysort(tmp, l, mid);
  res += mysort(tmp, mid + 1, r);
  long long t1 = l, t2 = mid + 1, t3 = 1;
  while (t1 <= mid && t2 <= r) {
    if (tmp[t1] <= tmp[t2])
      tmp2[t3++] = tmp[t1++], res += t2 - mid - 1;
    else
      tmp2[t3++] = tmp[t2++];
  }
  while (t1 <= mid) tmp2[t3++] = tmp[t1++], res += t2 - mid - 1;
  while (t2 <= r) tmp2[t3++] = tmp[t2++];
  for (int i = l; i <= r; i++) tmp[i] = tmp2[i - l + 1];
  return res;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
    sort(b + 1, b + 1 + m);
    mymid(a, b, pos, 1, n + 1, 1, m);
    int t1 = 1, t2 = 1, t3 = 1;
    while (t1 <= n + m) {
      while (pos[t2] <= t3 && t2 <= m) c[t1++] = b[t2++];
      if (t1 > n + m) break;
      c[t1++] = a[t3++];
    }
    printf("%lld\n", mysort(c, 1, n + m));
  }
  return 0;
}
