#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
namespace SGT {
int tag[N << 2], mn[N << 2];
inline void up(int rt) { mn[rt] = min(mn[(rt << 1)], mn[(rt << 1 | 1)]); }
inline void cov(int rt, int v) { mn[rt] += v, tag[rt] += v; }
inline void dn(int rt) {
  if (!tag[rt]) return;
  cov((rt << 1), tag[rt]), cov((rt << 1 | 1), tag[rt]);
  tag[rt] = 0;
}
void build(int rt, int l, int r) {
  tag[rt] = 0;
  if (l == r) return mn[rt] = l, void();
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid), build((rt << 1 | 1), mid + 1, r);
  up(rt);
}
void update(int rt, int l, int r, int ql, int qr, int v) {
  if (ql <= l && r <= qr) return cov(rt, v);
  dn(rt);
  int mid = (l + r) >> 1;
  if (ql <= mid) update((rt << 1), l, mid, ql, qr, v);
  if (mid < qr) update((rt << 1 | 1), mid + 1, r, ql, qr, v);
  up(rt);
}
}  // namespace SGT
pair<int, int> a[N];
int n, m, b[N], tmp[N];
namespace Fenwick {
int c[N];
inline void clear() { fill(c + 1, c + n + 1, 0); }
inline void upd(int x) {
  for (; x; x -= (x & (-x))) ++c[x];
}
inline int ask(int x) {
  int res = 0;
  for (; x <= n; x += (x & (-x))) res += c[x];
  return res;
}
}  // namespace Fenwick
using namespace Fenwick;
int main() {
  int T;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d%d", &n, &m);
    long long ans = 0;
    for (int i = 1; i <= n; ++i)
      scanf("%d", tmp + i), a[i] = make_pair(tmp[i], i);
    sort(tmp + 1, tmp + n + 1);
    int tot = unique(tmp + 1, tmp + n + 1) - tmp - 1;
    for (int i = 1; i <= n; ++i) {
      int now = lower_bound(tmp + 1, tmp + tot + 1, a[i].first) - tmp;
      ans += ask(now + 1);
      upd(now);
    }
    clear();
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= m; ++i) scanf("%d", b + i);
    sort(b + 1, b + m + 1);
    SGT::build(1, 0, n);
    int p1 = 1, p2 = 1;
    for (int i = 1; i <= m; ++i) {
      while (p1 <= n && a[p1].first < b[i])
        SGT::update(1, 0, n, 0, a[p1++].second - 1, 1);
      while (p2 <= n && a[p2].first <= b[i])
        SGT::update(1, 0, n, a[p2++].second, n, -1);
      ans += SGT::mn[1];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
namespace SGT {
int tag[N << 2], mn[N << 2];
inline void up(int rt) { mn[rt] = min(mn[(rt << 1)], mn[(rt << 1 | 1)]); }
inline void cov(int rt, int v) { mn[rt] += v, tag[rt] += v; }
inline void dn(int rt) {
  if (!tag[rt]) return;
  cov((rt << 1), tag[rt]), cov((rt << 1 | 1), tag[rt]);
  tag[rt] = 0;
}
void build(int rt, int l, int r) {
  tag[rt] = 0;
  if (l == r) return mn[rt] = l, void();
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid), build((rt << 1 | 1), mid + 1, r);
  up(rt);
}
void update(int rt, int l, int r, int ql, int qr, int v) {
  if (ql <= l && r <= qr) return cov(rt, v);
  dn(rt);
  int mid = (l + r) >> 1;
  if (ql <= mid) update((rt << 1), l, mid, ql, qr, v);
  if (mid < qr) update((rt << 1 | 1), mid + 1, r, ql, qr, v);
  up(rt);
}
}  // namespace SGT
pair<int, int> a[N];
int n, m, b[N], tmp[N];
namespace Fenwick {
int c[N];
inline void clear() { fill(c + 1, c + n + 1, 0); }
inline void upd(int x) {
  for (; x; x -= (x & (-x))) ++c[x];
}
inline int ask(int x) {
  int res = 0;
  for (; x <= n; x += (x & (-x))) res += c[x];
  return res;
}
}  // namespace Fenwick
using namespace Fenwick;
int main() {
  int T;
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d%d", &n, &m);
    long long ans = 0;
    for (int i = 1; i <= n; ++i)
      scanf("%d", tmp + i), a[i] = make_pair(tmp[i], i);
    sort(tmp + 1, tmp + n + 1);
    int tot = unique(tmp + 1, tmp + n + 1) - tmp - 1;
    for (int i = 1; i <= n; ++i) {
      int now = lower_bound(tmp + 1, tmp + tot + 1, a[i].first) - tmp;
      ans += ask(now + 1);
      upd(now);
    }
    clear();
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= m; ++i) scanf("%d", b + i);
    sort(b + 1, b + m + 1);
    SGT::build(1, 0, n);
    int p1 = 1, p2 = 1;
    for (int i = 1; i <= m; ++i) {
      while (p1 <= n && a[p1].first < b[i])
        SGT::update(1, 0, n, 0, a[p1++].second - 1, 1);
      while (p2 <= n && a[p2].first <= b[i])
        SGT::update(1, 0, n, a[p2++].second, n, -1);
      ans += SGT::mn[1];
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```