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;
using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;
int Bit(int mask, int b) { return (mask >> b) & 1; }
template <class T>
bool ckmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool ckmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
const ll INF = 2e18;
struct SegTree {
  struct Node {
    ll mn;
    ll upd;
    Node() : mn(INF), upd(0) {}
    void combine(const Node &a, const Node &b) { mn = min(a.mn, b.mn) + upd; }
    void check(ll x) {
      mn = x;
      upd = 0;
    }
    void update(ll x) {
      upd += x;
      mn += x;
    }
  };
  vector<Node> t;
  int n;
  void build(int sz) {
    n = 2;
    while (n < sz) n *= 2;
    t.resize(n * 2 + 5);
    for (int i = 0; i < (sz); ++i) {
      t[i + n].check(i);
    }
    for (int i = n - 1; i >= 1; --i) {
      t[i].combine(t[i << 1], t[i << 1 | 1]);
    }
  }
  void upd(int v, int l, int r, int lq, int rq, ll x) {
    if (lq >= r || l >= rq) return;
    if (lq <= l && r <= rq) {
      t[v].update(x);
      return;
    }
    upd(v << 1, l, (l + r) >> 1, lq, rq, x);
    upd(v << 1 | 1, (l + r) >> 1, r, lq, rq, x);
    t[v].combine(t[v << 1], t[v << 1 | 1]);
  }
  void upd(int l, int r, ll x) { upd(1, 0, n, l, r, x); }
  Node get(int v, int l, int r, int lq, int rq) {
    Node result;
    if (l >= rq || lq >= r) return result;
    if (lq <= l && r <= rq) {
      result = t[v];
      return result;
    }
    auto L = get(v << 1, l, (l + r) >> 1, lq, rq);
    auto R = get(v << 1 | 1, (l + r) >> 1, r, lq, rq);
    result.combine(L, R);
    result.update(t[v].upd);
    return result;
  }
  Node get(int lq, int rq) { return get(1, 0, n, lq, rq); }
};
struct fenwick {
  int n{};
  vector<ll> fenw{};
  void build(int k) {
    n = k;
    fenw.resize(n);
  }
  void upd(int i, ll x) {
    for (; i < n; i = i | (i + 1)) fenw[i] += x;
  }
  ll get(int i) {
    ll res = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) res += fenw[i];
    return res;
  }
  ll get(int l, int r) {
    if (l > r) return 0;
    return get(r) - get(l - 1);
  }
};
void solve() {
  int n, m;
  cin >> n >> m;
  vi a(n), b(m);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  for (int i = 0; i < (m); ++i) cin >> b[i];
  sort((b).begin(), (b).end());
  SegTree st;
  st.build(n + 1);
  int l = 0, r = 0;
  vector<pair<int, int>> c(n);
  for (int i = 0; i < (n); ++i) c[i] = {a[i], i};
  sort((c).begin(), (c).end());
  ll ans = 0;
  for (int i = 0; i < (m); ++i) {
    while (r < n && c[r].first <= b[i]) {
      st.upd(c[r].second + 1, n + 1, -1);
      r++;
    }
    while (l < n && c[l].first < b[i]) {
      st.upd(c[l].second + 1, n + 1, -1);
      l++;
    }
    ll cur = l;
    cur += st.get(0, n + 1).mn;
    ans += cur;
  }
  ll ans2 = 0;
  fenwick f;
  f.build(n);
  for (int i = 0, j = 0; i < n; i = j) {
    while (j < n && c[j].first == c[i].first) j++;
    for (int e = i; e < j; ++e) {
      ans2 += f.get(c[e].second + 1, n - 1);
    }
    for (int e = i; e < j; ++e) {
      f.upd(c[e].second, 1);
    }
  }
  cout << ans + ans2 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 0; i < (t); ++i) {
    solve();
  }
  return 0;
}
