#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 200005;
int n, m;
int maps[MAXN];
long long sum[MAXN][2];
pair<long long, long long> node[MAXN];
inline long long lowbit(long long x) { return x & (-x); }
void add(long long id, long long x) {
  while (id <= m) {
    sum[id][0]++;
    sum[id][1] += x;
    id += lowbit(id);
  }
}
long long query(int id, int op) {
  long long ans = 0;
  while (id > 0) {
    ans += sum[id][op];
    id -= lowbit(id);
  }
  return ans;
}
void Solve() {
  sort(node + 1, node + n + 1);
  sort(maps + 1, maps + n + 1);
  m = unique(maps + 1, maps + n + 1) - (maps + 1);
  long long ans = 0, id = -1;
  for (int i = 1; i <= n; ++i) {
    id = lower_bound(maps + 1, maps + m + 1, node[i].second) - maps;
    ans += node[i].first * query(id, 0) - query(id, 1);
    add(id, node[i].first);
  }
  cout << ans << endl;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> node[i].first;
  for (int i = 1; i <= n; ++i) {
    cin >> node[i].second;
    maps[i] = node[i].second;
  }
  Solve();
  return 0;
}

### Prompt

Create a solution in CPP for 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 double PI = acos(-1.0);
const double EPS = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 200005;
int n, m;
int maps[MAXN];
long long sum[MAXN][2];
pair<long long, long long> node[MAXN];
inline long long lowbit(long long x) { return x & (-x); }
void add(long long id, long long x) {
  while (id <= m) {
    sum[id][0]++;
    sum[id][1] += x;
    id += lowbit(id);
  }
}
long long query(int id, int op) {
  long long ans = 0;
  while (id > 0) {
    ans += sum[id][op];
    id -= lowbit(id);
  }
  return ans;
}
void Solve() {
  sort(node + 1, node + n + 1);
  sort(maps + 1, maps + n + 1);
  m = unique(maps + 1, maps + n + 1) - (maps + 1);
  long long ans = 0, id = -1;
  for (int i = 1; i <= n; ++i) {
    id = lower_bound(maps + 1, maps + m + 1, node[i].second) - maps;
    ans += node[i].first * query(id, 0) - query(id, 1);
    add(id, node[i].first);
  }
  cout << ans << endl;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> node[i].first;
  for (int i = 1; i <= n; ++i) {
    cin >> node[i].second;
    maps[i] = node[i].second;
  }
  Solve();
  return 0;
}

```