#include <bits/stdc++.h>
using namespace std;
template <class T>
class BIT {
 private:
  vector<T> dat;
  int n;

 public:
  BIT(int _n) : dat(_n + 1), n(_n) {}
  T sum(int i) {
    T s = 0;
    while (i > 0) s += dat[i], i -= i & -i;
    return s;
  }
  T sum(int l, int r) { return l > r ? 0 : sum(r) - sum(l - 1); }
  void add(int i, T x) {
    while (i <= n) dat[i] += x, i += i & -i;
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<pair<int64_t, int64_t>> P(N);
  for (auto &p : P) cin >> p.second;
  for (auto &p : P) cin >> p.first;
  sort(P.begin(), P.end());
  vector<int> vs;
  for (auto &p : P) vs.push_back(p.first);
  P[0].first = 1;
  for (int i = 1; i < N; ++i) {
    P[i].first = P[i - 1].first + (vs[i - 1] != vs[i]);
  }
  for (auto &p : P) swap(p.first, p.second);
  sort(P.begin(), P.end(), greater<pair<int, int>>());
  int64_t ans = 0LL;
  BIT<int64_t> bit(N), cnt(N);
  for (auto &p : P) {
    int x, v;
    tie(x, v) = p;
    ans += bit.sum(v, N) - x * cnt.sum(v, N);
    bit.add(v, x);
    cnt.add(v, 1);
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
template <class T>
class BIT {
 private:
  vector<T> dat;
  int n;

 public:
  BIT(int _n) : dat(_n + 1), n(_n) {}
  T sum(int i) {
    T s = 0;
    while (i > 0) s += dat[i], i -= i & -i;
    return s;
  }
  T sum(int l, int r) { return l > r ? 0 : sum(r) - sum(l - 1); }
  void add(int i, T x) {
    while (i <= n) dat[i] += x, i += i & -i;
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<pair<int64_t, int64_t>> P(N);
  for (auto &p : P) cin >> p.second;
  for (auto &p : P) cin >> p.first;
  sort(P.begin(), P.end());
  vector<int> vs;
  for (auto &p : P) vs.push_back(p.first);
  P[0].first = 1;
  for (int i = 1; i < N; ++i) {
    P[i].first = P[i - 1].first + (vs[i - 1] != vs[i]);
  }
  for (auto &p : P) swap(p.first, p.second);
  sort(P.begin(), P.end(), greater<pair<int, int>>());
  int64_t ans = 0LL;
  BIT<int64_t> bit(N), cnt(N);
  for (auto &p : P) {
    int x, v;
    tie(x, v) = p;
    ans += bit.sum(v, N) - x * cnt.sum(v, N);
    bit.add(v, x);
    cnt.add(v, 1);
  }
  cout << ans << '\n';
  return 0;
}

```