#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 20;
int n;
pair<int, int> p[N];
struct node {
  long long sum, count;
  node(long long s = 0, long long c = 0) : sum(s), count(c) {}
} bit[N];
void update(int x, int delta) {
  for (; x <= n; x += x & -x) {
    auto &[sum, count] = bit[x];
    sum += delta;
    count++;
  }
}
node query(int x) {
  auto ans = node();
  for (; x > 0; x -= x & -x) {
    auto &[sum, count] = bit[x];
    ans.sum += sum;
    ans.count += count;
  }
  return ans;
}
int compress(vector<int> &v) {
  if (v.empty()) return 0;
  vector<int> s = v;
  sort(s.begin(), s.end());
  s.resize(unique(s.begin(), s.end()) - s.begin());
  int mx = 1;
  for (int &x : v) {
    x = lower_bound(s.begin(), s.end(), x) - s.begin() + 1;
    mx = max(mx, x);
  }
  return mx;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) cin >> p[i].second;
  sort(p, p + n);
  vector<int> vec(n);
  for (int i = 0; i < n; i++) vec[i] = p[i].second;
  compress(vec);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    auto [sum, cnt] = query(vec[i]);
    ans += cnt * p[i].first - sum;
    update(vec[i], p[i].first);
  }
  cout << ans;
}

### Prompt

Please formulate 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 + 20;
int n;
pair<int, int> p[N];
struct node {
  long long sum, count;
  node(long long s = 0, long long c = 0) : sum(s), count(c) {}
} bit[N];
void update(int x, int delta) {
  for (; x <= n; x += x & -x) {
    auto &[sum, count] = bit[x];
    sum += delta;
    count++;
  }
}
node query(int x) {
  auto ans = node();
  for (; x > 0; x -= x & -x) {
    auto &[sum, count] = bit[x];
    ans.sum += sum;
    ans.count += count;
  }
  return ans;
}
int compress(vector<int> &v) {
  if (v.empty()) return 0;
  vector<int> s = v;
  sort(s.begin(), s.end());
  s.resize(unique(s.begin(), s.end()) - s.begin());
  int mx = 1;
  for (int &x : v) {
    x = lower_bound(s.begin(), s.end(), x) - s.begin() + 1;
    mx = max(mx, x);
  }
  return mx;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].first;
  for (int i = 0; i < n; i++) cin >> p[i].second;
  sort(p, p + n);
  vector<int> vec(n);
  for (int i = 0; i < n; i++) vec[i] = p[i].second;
  compress(vec);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    auto [sum, cnt] = query(vec[i]);
    ans += cnt * p[i].first - sum;
    update(vec[i], p[i].first);
  }
  cout << ans;
}

```