#include <bits/stdc++.h>
using namespace std;
int const maxn = 1e6 + 5;
int a[maxn], b[maxn];
pair<int, int> c[maxn];
int all_elem[2 * maxn];
int cur, inf = 1e9 + 7;
int f[2 * maxn];
int psh[8 * maxn];
pair<int, int> imin[4 * maxn];
inline pair<int, int> unions(pair<int, int> c, pair<int, int> d) {
  if (c.first < d.first || (c.first == d.first && c.second > d.second))
    return c;
  return d;
}
void build(int i, int l, int r) {
  psh[i] = 0;
  if (r - l == 1)
    imin[i] = {0, l};
  else {
    int m = (r + l) / 2;
    build(2 * i + 1, l, m);
    build(2 * i + 2, m, r);
    imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
  }
}
inline void push(int i) {
  imin[i].first += psh[i];
  psh[2 * i + 1] += psh[i];
  psh[2 * i + 2] += psh[i];
  psh[i] = 0;
}
void adds(int i, int l, int r, int lq, int rq, int x) {
  push(i);
  if (lq >= r || l >= rq) return;
  if (lq <= l && r <= rq) {
    psh[i] += x;
    push(i);
    return;
  }
  int m = (r + l) / 2;
  adds(2 * i + 1, l, m, lq, rq, x);
  adds(2 * i + 2, m, r, lq, rq, x);
  imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
}
pair<int, int> get(int i, int l, int r, int lq, int rq) {
  push(i);
  if (lq >= r || l >= rq) return {inf, inf};
  if (lq <= l && r <= rq) return imin[i];
  int m = (r + l) / 2;
  return unions(get(2 * i + 1, l, m, lq, rq), get(2 * i + 2, m, r, lq, rq));
}
inline void add(int i) {
  while (i <= cur) {
    f[i]++;
    i = (i | (i + 1));
  }
}
inline int get(int i) {
  int ans = 0;
  while (i > 0) {
    ans += f[i];
    i = (i & (i + 1)) - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cur = 0;
    cin >> n >> m;
    build(0, 0, n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= m; ++i) cin >> b[i];
    sort(b + 1, b + m + 1);
    for (int i = 1; i <= n; ++i) {
      all_elem[++cur] = a[i];
    }
    for (int i = 1; i <= m; ++i) {
      all_elem[++cur] = b[i];
    }
    sort(all_elem + 1, all_elem + cur + 1);
    for (int i = 1; i <= n; ++i) {
      a[i] = lower_bound(all_elem + 1, all_elem + cur + 1, a[i]) - all_elem;
    }
    for (int i = 1; i <= m; ++i) {
      b[i] = lower_bound(all_elem + 1, all_elem + cur + 1, b[i]) - all_elem;
    }
    long long ans = 0;
    for (int i = 0; i <= cur; ++i) f[i] = 0;
    for (int i = n; i >= 1; --i) {
      ans += get(a[i]);
      add(a[i] + 1);
    }
    for (int i = 1; i <= n; ++i) c[i] = {a[i], i};
    sort(c + 1, c + n + 1);
    int pos = n;
    for (int i = 1; i <= n; ++i) {
      adds(0, 0, n + 1, 0, i, 1);
    }
    int lptr = n, rptr = n;
    for (int i = m; i >= 1; --i) {
      if (b[i] != b[i + 1]) {
        while (rptr > lptr) {
          adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
          rptr--;
        }
      }
      while (rptr > 0 && b[i] < c[rptr].first) {
        adds(0, 0, n + 1, 0, c[rptr].second, -1);
        adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
        rptr--;
      }
      lptr = min(lptr, rptr);
      while (lptr > 0 && b[i] == c[lptr].first) {
        adds(0, 0, n + 1, 0, c[lptr].second, -1);
        lptr--;
      }
      pair<int, int> best = get(0, 0, n + 1, 0, pos + 1);
      ans += best.first;
      pos = best.second;
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int const maxn = 1e6 + 5;
int a[maxn], b[maxn];
pair<int, int> c[maxn];
int all_elem[2 * maxn];
int cur, inf = 1e9 + 7;
int f[2 * maxn];
int psh[8 * maxn];
pair<int, int> imin[4 * maxn];
inline pair<int, int> unions(pair<int, int> c, pair<int, int> d) {
  if (c.first < d.first || (c.first == d.first && c.second > d.second))
    return c;
  return d;
}
void build(int i, int l, int r) {
  psh[i] = 0;
  if (r - l == 1)
    imin[i] = {0, l};
  else {
    int m = (r + l) / 2;
    build(2 * i + 1, l, m);
    build(2 * i + 2, m, r);
    imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
  }
}
inline void push(int i) {
  imin[i].first += psh[i];
  psh[2 * i + 1] += psh[i];
  psh[2 * i + 2] += psh[i];
  psh[i] = 0;
}
void adds(int i, int l, int r, int lq, int rq, int x) {
  push(i);
  if (lq >= r || l >= rq) return;
  if (lq <= l && r <= rq) {
    psh[i] += x;
    push(i);
    return;
  }
  int m = (r + l) / 2;
  adds(2 * i + 1, l, m, lq, rq, x);
  adds(2 * i + 2, m, r, lq, rq, x);
  imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
}
pair<int, int> get(int i, int l, int r, int lq, int rq) {
  push(i);
  if (lq >= r || l >= rq) return {inf, inf};
  if (lq <= l && r <= rq) return imin[i];
  int m = (r + l) / 2;
  return unions(get(2 * i + 1, l, m, lq, rq), get(2 * i + 2, m, r, lq, rq));
}
inline void add(int i) {
  while (i <= cur) {
    f[i]++;
    i = (i | (i + 1));
  }
}
inline int get(int i) {
  int ans = 0;
  while (i > 0) {
    ans += f[i];
    i = (i & (i + 1)) - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cur = 0;
    cin >> n >> m;
    build(0, 0, n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= m; ++i) cin >> b[i];
    sort(b + 1, b + m + 1);
    for (int i = 1; i <= n; ++i) {
      all_elem[++cur] = a[i];
    }
    for (int i = 1; i <= m; ++i) {
      all_elem[++cur] = b[i];
    }
    sort(all_elem + 1, all_elem + cur + 1);
    for (int i = 1; i <= n; ++i) {
      a[i] = lower_bound(all_elem + 1, all_elem + cur + 1, a[i]) - all_elem;
    }
    for (int i = 1; i <= m; ++i) {
      b[i] = lower_bound(all_elem + 1, all_elem + cur + 1, b[i]) - all_elem;
    }
    long long ans = 0;
    for (int i = 0; i <= cur; ++i) f[i] = 0;
    for (int i = n; i >= 1; --i) {
      ans += get(a[i]);
      add(a[i] + 1);
    }
    for (int i = 1; i <= n; ++i) c[i] = {a[i], i};
    sort(c + 1, c + n + 1);
    int pos = n;
    for (int i = 1; i <= n; ++i) {
      adds(0, 0, n + 1, 0, i, 1);
    }
    int lptr = n, rptr = n;
    for (int i = m; i >= 1; --i) {
      if (b[i] != b[i + 1]) {
        while (rptr > lptr) {
          adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
          rptr--;
        }
      }
      while (rptr > 0 && b[i] < c[rptr].first) {
        adds(0, 0, n + 1, 0, c[rptr].second, -1);
        adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
        rptr--;
      }
      lptr = min(lptr, rptr);
      while (lptr > 0 && b[i] == c[lptr].first) {
        adds(0, 0, n + 1, 0, c[lptr].second, -1);
        lptr--;
      }
      pair<int, int> best = get(0, 0, n + 1, 0, pos + 1);
      ans += best.first;
      pos = best.second;
    }
    cout << ans << '\n';
  }
  return 0;
}

```