#include <bits/stdc++.h>
using namespace std;
int n;
long long bit[200005][2];
void update(int x, long long val) {
  for (int i = x; i <= n; i += (i & -i)) bit[i][0] += val;
  for (int i = x; i <= n; i += (i & -i)) bit[i][1] += 1;
}
pair<long long, long long> query(int x) {
  long long ret1 = 0;
  long long ret2 = 0;
  for (int i = x; i > 0; i -= (i & -i)) ret1 += bit[i][0];
  for (int i = x; i > 0; i -= (i & -i)) ret2 += bit[i][1];
  return {ret1, ret2};
}
int main() {
  cin >> n;
  vector<long long> vel, tmp;
  vector<pair<long long, int> > pos;
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    pos.push_back({x, i});
  }
  sort(pos.begin(), pos.end());
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    vel.push_back(x);
    tmp.push_back(x);
  }
  sort(tmp.begin(), tmp.end());
  for (int i = 0; i < n; i++) {
    mp[tmp[i]] = i + 1;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int posnow = pos[i].second;
    long long xnow = pos[i].first;
    int now = mp[vel[posnow]];
    pair<long long, long long> q = query(now);
    long long val = xnow * q.second - q.first;
    ans += val;
    update(now, xnow);
  }
  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;
int n;
long long bit[200005][2];
void update(int x, long long val) {
  for (int i = x; i <= n; i += (i & -i)) bit[i][0] += val;
  for (int i = x; i <= n; i += (i & -i)) bit[i][1] += 1;
}
pair<long long, long long> query(int x) {
  long long ret1 = 0;
  long long ret2 = 0;
  for (int i = x; i > 0; i -= (i & -i)) ret1 += bit[i][0];
  for (int i = x; i > 0; i -= (i & -i)) ret2 += bit[i][1];
  return {ret1, ret2};
}
int main() {
  cin >> n;
  vector<long long> vel, tmp;
  vector<pair<long long, int> > pos;
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    pos.push_back({x, i});
  }
  sort(pos.begin(), pos.end());
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    vel.push_back(x);
    tmp.push_back(x);
  }
  sort(tmp.begin(), tmp.end());
  for (int i = 0; i < n; i++) {
    mp[tmp[i]] = i + 1;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int posnow = pos[i].second;
    long long xnow = pos[i].first;
    int now = mp[vel[posnow]];
    pair<long long, long long> q = query(now);
    long long val = xnow * q.second - q.first;
    ans += val;
    update(now, xnow);
  }
  cout << ans << endl;
  return 0;
}

```