#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int MOD = 1000000007;
const long long INFLL = 1e18;
const int INF = 1e9;
const int NMAX = 2e5 + 5;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int n;
array<int, 2> a[NMAX];
struct BIT {
  long long aib[NMAX];
  void update(int pos, int val) {
    for (int i = pos + 1; i <= n; i += (i & (-i))) {
      aib[i] += val;
    }
  }
  long long query(int pos) {
    long long sum = 0;
    for (int i = pos; i >= 1; i -= (i & (-i))) {
      sum += aib[i];
    }
    return sum;
  }
} f0, f1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i][0];
  }
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i][1];
    v.push_back(a[i][1]);
  }
  long long ans = 0;
  sort(v.begin(), v.end());
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    int pos = lower_bound(v.begin(), v.end(), a[i][1]) - v.begin();
    ans += f0.query(pos + 1) * a[i][0] - f1.query(pos + 1);
    f0.update(pos, 1);
    f1.update(pos, a[i][0]);
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in CPP for 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 c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const int MOD = 1000000007;
const long long INFLL = 1e18;
const int INF = 1e9;
const int NMAX = 2e5 + 5;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int n;
array<int, 2> a[NMAX];
struct BIT {
  long long aib[NMAX];
  void update(int pos, int val) {
    for (int i = pos + 1; i <= n; i += (i & (-i))) {
      aib[i] += val;
    }
  }
  long long query(int pos) {
    long long sum = 0;
    for (int i = pos; i >= 1; i -= (i & (-i))) {
      sum += aib[i];
    }
    return sum;
  }
} f0, f1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i][0];
  }
  vector<int> v;
  for (int i = 0; i < n; ++i) {
    cin >> a[i][1];
    v.push_back(a[i][1]);
  }
  long long ans = 0;
  sort(v.begin(), v.end());
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    int pos = lower_bound(v.begin(), v.end(), a[i][1]) - v.begin();
    ans += f0.query(pos + 1) * a[i][0] - f1.query(pos + 1);
    f0.update(pos, 1);
    f1.update(pos, a[i][0]);
  }
  cout << ans;
  return 0;
}

```