#include <bits/stdc++.h>
using namespace std;
struct FT {
  vector<long long> f;
  FT(long long n) { f.resize(n + 1, 0); }
  long long sum(long long r) {
    long long res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      res += f[r];
    }
    return res;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
  void upd(long long i, long long x) {
    for (; i < (long long)f.size(); i = (i | (i + 1))) f[i] += x;
  }
};
map<long long, long long> f;
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second, f[a[i].second] = 0;
  long long s = 0;
  for (auto u : f) f[u.first] = s++;
  sort(a.begin(), a.end());
  FT ft1(s + 1), ft2(s + 1);
  long long ans = 0;
  for (auto u : a) {
    long long x = u.first, y = u.second;
    long long v = f[y];
    long long cnt = ft1.sum(0, v);
    long long sum = ft2.sum(0, v);
    ans += x * cnt - sum;
    ft1.upd(v, 1);
    ft2.upd(v, x);
  }
  cout << ans << '\n';
}

### Prompt

Generate 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;
struct FT {
  vector<long long> f;
  FT(long long n) { f.resize(n + 1, 0); }
  long long sum(long long r) {
    long long res = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      res += f[r];
    }
    return res;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
  void upd(long long i, long long x) {
    for (; i < (long long)f.size(); i = (i | (i + 1))) f[i] += x;
  }
};
map<long long, long long> f;
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i].first;
  for (long long i = 0; i < n; i++) cin >> a[i].second, f[a[i].second] = 0;
  long long s = 0;
  for (auto u : f) f[u.first] = s++;
  sort(a.begin(), a.end());
  FT ft1(s + 1), ft2(s + 1);
  long long ans = 0;
  for (auto u : a) {
    long long x = u.first, y = u.second;
    long long v = f[y];
    long long cnt = ft1.sum(0, v);
    long long sum = ft2.sum(0, v);
    ans += x * cnt - sum;
    ft1.upd(v, 1);
    ft2.upd(v, x);
  }
  cout << ans << '\n';
}

```