#include <bits/stdc++.h>
using namespace std;
int N;
pair<signed long long, signed long long> P[202020];
pair<signed long long, signed long long> Q[202020];
template <class V, int ME>
class BIT {
 public:
  V bit[1 << ME];
  V operator()(int e) {
    if (e < 0) return 0;
    V s = 0;
    e++;
    while (e) s += bit[e - 1], e -= e & -e;
    return s;
  }
  void add(int e, V v) {
    e++;
    while (e <= 1 << ME) bit[e - 1] += v, e += e & -e;
  }
};
BIT<signed long long, 20> bt, bt2;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < (N); i++) cin >> P[i].first;
  for (i = 0; i < (N); i++) cin >> P[i].second;
  sort(P, P + N);
  signed long long ret = 0;
  for (i = 0; i < (N); i++) Q[i] = {P[i].second, i};
  sort(Q, Q + N);
  for (i = 0; i < (N); i++) {
    ret += bt(Q[i].second) * P[Q[i].second].first;
    bt.add(Q[i].second, 1);
  }
  reverse(Q, Q + N);
  for (i = 0; i < (N); i++) {
    ret -= (bt2(N) - bt2(Q[i].second)) * P[Q[i].second].first;
    bt2.add(Q[i].second, 1);
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
int N;
pair<signed long long, signed long long> P[202020];
pair<signed long long, signed long long> Q[202020];
template <class V, int ME>
class BIT {
 public:
  V bit[1 << ME];
  V operator()(int e) {
    if (e < 0) return 0;
    V s = 0;
    e++;
    while (e) s += bit[e - 1], e -= e & -e;
    return s;
  }
  void add(int e, V v) {
    e++;
    while (e <= 1 << ME) bit[e - 1] += v, e += e & -e;
  }
};
BIT<signed long long, 20> bt, bt2;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < (N); i++) cin >> P[i].first;
  for (i = 0; i < (N); i++) cin >> P[i].second;
  sort(P, P + N);
  signed long long ret = 0;
  for (i = 0; i < (N); i++) Q[i] = {P[i].second, i};
  sort(Q, Q + N);
  for (i = 0; i < (N); i++) {
    ret += bt(Q[i].second) * P[Q[i].second].first;
    bt.add(Q[i].second, 1);
  }
  reverse(Q, Q + N);
  for (i = 0; i < (N); i++) {
    ret -= (bt2(N) - bt2(Q[i].second)) * P[Q[i].second].first;
    bt2.add(Q[i].second, 1);
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

```