#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("no-stack-protector")
using namespace std;
const int inf = 1e9 + 7;
const int max_n = 2e5 + 1;
struct fenwick {
  vector<int> sum;
  fenwick(int n) {
    sum.clear();
    sum.resize(n, 0);
  }
  int f(int v) { return (v & (v + 1)); }
  int h(int v) { return (v | (v + 1)); }
  void add(int v) {
    while (v < sum.size()) {
      sum[v]++;
      v = h(v);
    }
  }
  int get(int r) {
    int ans = 0;
    while (r >= 0) {
      ans += sum[r];
      r = f(r) - 1;
    }
    return ans;
  }
  int get(int l, int r) { return get(r) - get(l - 1); }
};
int n;
pair<int, int> a[max_n];
int sq[max_n];
void scan() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  map<int, vector<int>> temp;
  for (int i = 0; i < n; i++) temp[a[i].second].emplace_back(i);
  int cnt = 0;
  for (auto i : temp) {
    for (int j : i.second) {
      sq[j] = cnt;
      cnt++;
    }
  }
}
void solve() {
  scan();
  long long ans = 0;
  fenwick one = fenwick(n);
  for (int i = 0; i < n; i++) {
    ans += (long long)one.get(sq[i]) * a[i].first;
    one.add(sq[i]);
  }
  fenwick two = fenwick(n);
  for (int i = n - 1; i >= 0; i--) {
    ans -= (long long)two.get(sq[i], n - 1) * a[i].first;
    two.add(sq[i]);
  }
  cout << ans;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  solve();
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma GCC optimize("O3")
#pragma GCC optimize("no-stack-protector")
using namespace std;
const int inf = 1e9 + 7;
const int max_n = 2e5 + 1;
struct fenwick {
  vector<int> sum;
  fenwick(int n) {
    sum.clear();
    sum.resize(n, 0);
  }
  int f(int v) { return (v & (v + 1)); }
  int h(int v) { return (v | (v + 1)); }
  void add(int v) {
    while (v < sum.size()) {
      sum[v]++;
      v = h(v);
    }
  }
  int get(int r) {
    int ans = 0;
    while (r >= 0) {
      ans += sum[r];
      r = f(r) - 1;
    }
    return ans;
  }
  int get(int l, int r) { return get(r) - get(l - 1); }
};
int n;
pair<int, int> a[max_n];
int sq[max_n];
void scan() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i].first;
  for (int i = 0; i < n; i++) cin >> a[i].second;
  sort(a, a + n);
  map<int, vector<int>> temp;
  for (int i = 0; i < n; i++) temp[a[i].second].emplace_back(i);
  int cnt = 0;
  for (auto i : temp) {
    for (int j : i.second) {
      sq[j] = cnt;
      cnt++;
    }
  }
}
void solve() {
  scan();
  long long ans = 0;
  fenwick one = fenwick(n);
  for (int i = 0; i < n; i++) {
    ans += (long long)one.get(sq[i]) * a[i].first;
    one.add(sq[i]);
  }
  fenwick two = fenwick(n);
  for (int i = n - 1; i >= 0; i--) {
    ans -= (long long)two.get(sq[i], n - 1) * a[i].first;
    two.add(sq[i]);
  }
  cout << ans;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  solve();
}

```