#include <bits/stdc++.h>
using namespace std;
const long long mxn = 1e6 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, -1, +1};
pair<int, int> p[mxn];
map<long long, long long> f, s;
int n;
int a[mxn], ls[mxn];
long long ans;
void upd1(int x, int val) {
  while (x <= (int)1e8) {
    f[x] += val;
    x = (x | (x + 1));
  }
}
void upd2(int x) {
  while (x <= (int)1e8) {
    s[x]++;
    x = (x | (x + 1));
  }
}
long long sum1(int x) {
  long long res = 0;
  while (x > 0) {
    res += f[x];
    x = (x & (x + 1)) - 1;
  }
  return res;
}
int sum2(int x) {
  int res = 0;
  while (x > 0) {
    res += s[x];
    x = (x & (x + 1)) - 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    ls[i] = p[i].second;
  }
  sort(p + 1, p + n + 1);
  sort(ls + 1, ls + n + 1);
  for (int i = 1; i <= n; i++) {
    p[i].second = lower_bound(ls + 1, ls + n + 1, p[i].second) - ls;
  }
  for (int i = 1; i <= n; i++) {
    ans += (1LL * p[i].first * sum2(p[i].second)) - sum1(p[i].second);
    upd1(p[i].second, p[i].first);
    upd2(p[i].second);
  }
  cout << ans;
  return 0;
}

### 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;
const long long mxn = 1e6 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 7;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, -1, +1};
pair<int, int> p[mxn];
map<long long, long long> f, s;
int n;
int a[mxn], ls[mxn];
long long ans;
void upd1(int x, int val) {
  while (x <= (int)1e8) {
    f[x] += val;
    x = (x | (x + 1));
  }
}
void upd2(int x) {
  while (x <= (int)1e8) {
    s[x]++;
    x = (x | (x + 1));
  }
}
long long sum1(int x) {
  long long res = 0;
  while (x > 0) {
    res += f[x];
    x = (x & (x + 1)) - 1;
  }
  return res;
}
int sum2(int x) {
  int res = 0;
  while (x > 0) {
    res += s[x];
    x = (x & (x + 1)) - 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> p[i].second;
    ls[i] = p[i].second;
  }
  sort(p + 1, p + n + 1);
  sort(ls + 1, ls + n + 1);
  for (int i = 1; i <= n; i++) {
    p[i].second = lower_bound(ls + 1, ls + n + 1, p[i].second) - ls;
  }
  for (int i = 1; i <= n; i++) {
    ans += (1LL * p[i].first * sum2(p[i].second)) - sum1(p[i].second);
    upd1(p[i].second, p[i].first);
    upd2(p[i].second);
  }
  cout << ans;
  return 0;
}

```