#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100011;
const long long INF = (1 << 29) + 123;
const long long MOD = 1000000007;
const long double PI = 4 * atan((long double)1);
int fenw[2 * MAX_N];
int n;
void mod(int i, int v) {
  while (i <= n) {
    fenw[i] += v;
    i += i & -i;
  }
}
int query(int i) {
  int sum = 0;
  while (i > 0) {
    sum += fenw[i];
    i -= i & -i;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int> > p;
  int x[n], v[n];
  for (int i = 0; i < n; i++) cin >> x[i];
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    p.push_back(make_pair(v[i], x[i]));
  }
  unordered_map<int, int> m;
  int rev_map[n];
  sort(x, x + n);
  for (int i = 0; i < n; i++) {
    m[x[i]] = i + 1;
    rev_map[i] = x[i];
  }
  for (int i = 0; i < n; i++) {
    p[i].second = m[p[i].second];
  }
  sort(p.begin(), p.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int q = query(p[i].second);
    ans += 1LL * q * rev_map[p[i].second - 1];
    ans -=
        1LL * ((n - i - 1) - (p[i].second - 1 - q)) * rev_map[p[i].second - 1];
    mod(p[i].second, 1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 MAX_N = 100011;
const long long INF = (1 << 29) + 123;
const long long MOD = 1000000007;
const long double PI = 4 * atan((long double)1);
int fenw[2 * MAX_N];
int n;
void mod(int i, int v) {
  while (i <= n) {
    fenw[i] += v;
    i += i & -i;
  }
}
int query(int i) {
  int sum = 0;
  while (i > 0) {
    sum += fenw[i];
    i -= i & -i;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int> > p;
  int x[n], v[n];
  for (int i = 0; i < n; i++) cin >> x[i];
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    p.push_back(make_pair(v[i], x[i]));
  }
  unordered_map<int, int> m;
  int rev_map[n];
  sort(x, x + n);
  for (int i = 0; i < n; i++) {
    m[x[i]] = i + 1;
    rev_map[i] = x[i];
  }
  for (int i = 0; i < n; i++) {
    p[i].second = m[p[i].second];
  }
  sort(p.begin(), p.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int q = query(p[i].second);
    ans += 1LL * q * rev_map[p[i].second - 1];
    ans -=
        1LL * ((n - i - 1) - (p[i].second - 1 - q)) * rev_map[p[i].second - 1];
    mod(p[i].second, 1);
  }
  cout << ans << endl;
  return 0;
}

```