#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int bit[N];
void add(int p, int v) {
  for (p += 2; p < N; p += p & -p) bit[p] += v;
}
long long query(int p) {
  int r = 0;
  for (p += 2; p; p -= p & -p) r += bit[p];
  return r;
}
int n;
pair<int, int> p[N];
long long l[N], r[N];
int ql[N], qr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].second;
  for (int i = 1; i <= n; i++) cin >> p[i].first;
  sort(p + 1, p + n + 1);
  map<int, int> id;
  int cnt = 0, last = 0x3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    int v = p[i].first;
    if (v != last) id[v] = ++cnt, last = v;
    swap(p[i].first, p[i].second);
  }
  sort(p + 1, p + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x = p[i].first, v = id[p[i].second];
    ans += query(v) * x;
    add(v, 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n; i >= 1; i--) {
    long long x = p[i].first, v = id[p[i].second];
    ans -= (query(n) - query(v - 1)) * x;
    add(v, 1);
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create 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 int N = 2e5 + 5;
int bit[N];
void add(int p, int v) {
  for (p += 2; p < N; p += p & -p) bit[p] += v;
}
long long query(int p) {
  int r = 0;
  for (p += 2; p; p -= p & -p) r += bit[p];
  return r;
}
int n;
pair<int, int> p[N];
long long l[N], r[N];
int ql[N], qr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].second;
  for (int i = 1; i <= n; i++) cin >> p[i].first;
  sort(p + 1, p + n + 1);
  map<int, int> id;
  int cnt = 0, last = 0x3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    int v = p[i].first;
    if (v != last) id[v] = ++cnt, last = v;
    swap(p[i].first, p[i].second);
  }
  sort(p + 1, p + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x = p[i].first, v = id[p[i].second];
    ans += query(v) * x;
    add(v, 1);
  }
  memset(bit, 0, sizeof(bit));
  for (int i = n; i >= 1; i--) {
    long long x = p[i].first, v = id[p[i].second];
    ans -= (query(n) - query(v - 1)) * x;
    add(v, 1);
  }
  cout << ans << '\n';
}

```