#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
pair<int, int> p[N];
struct fenwick {
  long long fen[N];
  void add(int first, int d) {
    for (int i = first + 1; i < N; i += i & -i) {
      fen[i] += d;
    }
  }
  long long sum(int first) {
    long long ans = 0;
    for (int i = first + 1; i; i -= i & -i) {
      ans += fen[i];
    }
    return ans;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
} val, cnt;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].first;
  for (int i = 0; i < n; ++i) cin >> p[i].second;
  sort(p, p + n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) a[i] = p[i].second;
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  for (int i = 0; i < n; ++i) {
    p[i].second = lower_bound(a.begin(), a.end(), p[i].second) - a.begin();
  }
  long long ans = 0;
  for (int i = n - 1; ~i; --i) {
    val.add(p[i].second, p[i].first);
    cnt.add(p[i].second, 1);
    ans += val.sum(p[i].second, 2e5) - cnt.sum(p[i].second, 2e5) * p[i].first;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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 N = 2e5 + 7;
pair<int, int> p[N];
struct fenwick {
  long long fen[N];
  void add(int first, int d) {
    for (int i = first + 1; i < N; i += i & -i) {
      fen[i] += d;
    }
  }
  long long sum(int first) {
    long long ans = 0;
    for (int i = first + 1; i; i -= i & -i) {
      ans += fen[i];
    }
    return ans;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
} val, cnt;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].first;
  for (int i = 0; i < n; ++i) cin >> p[i].second;
  sort(p, p + n);
  vector<int> a(n);
  for (int i = 0; i < n; ++i) a[i] = p[i].second;
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  for (int i = 0; i < n; ++i) {
    p[i].second = lower_bound(a.begin(), a.end(), p[i].second) - a.begin();
  }
  long long ans = 0;
  for (int i = n - 1; ~i; --i) {
    val.add(p[i].second, p[i].first);
    cnt.add(p[i].second, 1);
    ans += val.sum(p[i].second, 2e5) - cnt.sum(p[i].second, 2e5) * p[i].first;
  }
  cout << ans << '\n';
  return 0;
}

```