#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n;
pair<int, int> a[maxn];
int m;
vector<int> v;
struct fenwick {
  long long val[maxn];
  void init() { fill(begin(val), end(val), 0); }
  void upd(int x, int k) {
    for (; x < maxn; x += x & -x) val[x] += k;
  }
  long long get(int x) {
    long long res = 0;
    for (; x > 0; x -= x & -x) res += val[x];
    return res;
  }
  long long get(int l, int r) {
    if (l > r) return 0;
    return get(r) - get(l - 1);
  }
} tree_tot, tree_cnt;
void read_input() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first;
  for (int i = 1; i <= n; ++i) cin >> a[i].second;
}
void init() {
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) v.push_back(a[i].second);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  m = v.size();
  for (int i = 1; i <= n; ++i)
    a[i].second = lower_bound(v.begin(), v.end(), a[i].second) - v.begin() + 1;
  tree_tot.init();
  tree_cnt.init();
  for (int i = 1; i <= n; ++i) {
    tree_tot.upd(a[i].second, a[i].first);
    tree_cnt.upd(a[i].second, 1);
  }
}
void solve() {
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    tree_tot.upd(a[i].second, -a[i].first);
    tree_cnt.upd(a[i].second, -1);
    ans += tree_tot.get(a[i].second, m) -
           a[i].first * 1LL * tree_cnt.get(a[i].second, m);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  read_input();
  init();
  solve();
}

### Prompt

Please create a solution in Cpp to the following problem:

There are n points on a coordinate axis OX. The i-th point is located at the integer point x_i and has a speed v_i. It is guaranteed that no two points occupy the same coordinate. All n points move with the constant speed, the coordinate of the i-th point at the moment t (t can be non-integer) is calculated as x_i + t ⋅ v_i.

Consider two points i and j. Let d(i, j) be the minimum possible distance between these two points over any possible moments of time (even non-integer). It means that if two points i and j coincide at some moment, the value d(i, j) will be 0.

Your task is to calculate the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of points.

The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^8), where x_i is the initial coordinate of the i-th point. It is guaranteed that all x_i are distinct.

The third line of the input contains n integers v_1, v_2, ..., v_n (-10^8 ≤ v_i ≤ 10^8), where v_i is the speed of the i-th point.

Output

Print one integer — the value ∑_{1 ≤ i < j ≤ n} d(i, j) (the sum of minimum distances over all pairs of points).

Examples

Input


3
1 3 2
-100 2 3


Output


3


Input


5
2 1 4 3 5
2 2 2 3 4


Output


19


Input


2
2 1
-3 0


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n;
pair<int, int> a[maxn];
int m;
vector<int> v;
struct fenwick {
  long long val[maxn];
  void init() { fill(begin(val), end(val), 0); }
  void upd(int x, int k) {
    for (; x < maxn; x += x & -x) val[x] += k;
  }
  long long get(int x) {
    long long res = 0;
    for (; x > 0; x -= x & -x) res += val[x];
    return res;
  }
  long long get(int l, int r) {
    if (l > r) return 0;
    return get(r) - get(l - 1);
  }
} tree_tot, tree_cnt;
void read_input() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first;
  for (int i = 1; i <= n; ++i) cin >> a[i].second;
}
void init() {
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) v.push_back(a[i].second);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  m = v.size();
  for (int i = 1; i <= n; ++i)
    a[i].second = lower_bound(v.begin(), v.end(), a[i].second) - v.begin() + 1;
  tree_tot.init();
  tree_cnt.init();
  for (int i = 1; i <= n; ++i) {
    tree_tot.upd(a[i].second, a[i].first);
    tree_cnt.upd(a[i].second, 1);
  }
}
void solve() {
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    tree_tot.upd(a[i].second, -a[i].first);
    tree_cnt.upd(a[i].second, -1);
    ans += tree_tot.get(a[i].second, m) -
           a[i].first * 1LL * tree_cnt.get(a[i].second, m);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  read_input();
  init();
  solve();
}

```