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;
const ll maxn = 2e6 + 10;
const ll inf = 0x3f3f3f3f;
ll a[maxn], b[maxn], p[maxn];
ll n, m;
map<ll, ll> dic;
ll bits[maxn];
ll lowbit(ll x) { return x & (-x); }
void update(ll x, ll pos) {
  while (pos < maxn) {
    bits[pos] += x;
    pos += lowbit(pos);
  }
}
ll query(ll pos) {
  ll res = 0;
  while (pos) {
    res += bits[pos];
    pos -= lowbit(pos);
  }
  return res;
}
void insert(ll lb, ll rb, ll la, ll ra) {
  if (lb > rb || la > ra) return;
  ll nowb = b[(lb + rb) / 2];
  ll less = 0;
  for (ll pos = la; pos < ra; ++pos) {
    if (a[pos] < nowb) ++less;
  }
  ll pick = la, ans = less;
  ll now = la;
  while (now < ra) {
    if (a[now] != nowb) {
      if (a[now] < nowb)
        --less;
      else
        ++less;
    }
    ++now;
    if (ans > less) {
      ans = less;
      pick = now;
    }
  }
  p[(lb + rb) / 2] = pick;
  insert(lb, (lb + rb) / 2 - 1, la, pick);
  insert((lb + rb) / 2 + 1, rb, pick, ra);
}
void solve() {
  dic.clear();
  cin >> n >> m;
  for (ll i = 1; i <= n; ++i) cin >> a[i];
  for (ll i = 1; i <= m; ++i) cin >> b[i];
  for (ll i = 1; i <= m + n + 2; ++i) bits[i] = 0;
  sort(b + 1, b + m + 1);
  insert(1, m, 1, n + 1);
  vector<ll> discrete;
  for (ll i = 1; i <= n; ++i) {
    discrete.emplace_back(a[i]);
  }
  for (ll i = 1; i <= m; ++i) {
    discrete.emplace_back(b[i]);
  }
  sort(discrete.begin(), discrete.end());
  discrete.erase(unique(discrete.begin(), discrete.end()), discrete.end());
  for (ll i = 0; i < discrete.size(); ++i) {
    dic[discrete[i]] = i + 1;
  }
  vector<ll> c;
  ll j = 1;
  for (ll i = 1; i <= n; ++i) {
    while (j <= m && p[j] == i) c.push_back(b[j]), ++j;
    c.push_back(a[i]);
  }
  while (j <= m) c.push_back(b[j]), ++j;
  ll ans = 0;
  for (ll i = 0; i < c.size(); ++i) {
    ll rk = dic[c[i]];
    ans += i - query(rk);
    update(1, rk);
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ll t;
  cin >> t;
  while (t--) {
    solve();
  }
}
