#include <bits/stdc++.h>
using namespace std;
template <typename C>
class fenwick {
  vector<C> t;
  int n;

 public:
  fenwick(int _n) : n(_n) { t.resize(n); }
  fenwick(const vector<C>& a) : n(a.size()) {
    t.resize(n);
    for (int i = 0; i < n; ++i) {
      upd(i, a[i]);
    }
  }
  C get(int k) {
    C res = 0;
    for (k; k >= 0; k = (k & (k + 1)) - 1) {
      res += t[k];
    }
    return res;
  }
  void upd(int k, C x) {
    for (k; k < n; k |= k + 1) {
      t[k] += x;
    }
  }
  C get(int l, int r) { return get(r) - get(l - 1); }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> x(n), v(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
  }
  map<int, int> mapchik;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    mapchik[v[i]] = 1;
  }
  int sz = 0;
  for (auto& [a, b] : mapchik) {
    b = sz++;
  }
  for (int i = 0; i < n; ++i) {
    v[i] = mapchik[v[i]];
  }
  vector<int> ord(n);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return x[i] < x[j]; });
  long long ans = 0;
  fenwick<int> cnt(sz);
  fenwick<long long> sum(sz);
  for (int i : ord) {
    ans += 1ll * cnt.get(0, v[i]) * x[i] - sum.get(0, v[i]);
    cnt.upd(v[i], 1);
    sum.upd(v[i], x[i]);
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <typename C>
class fenwick {
  vector<C> t;
  int n;

 public:
  fenwick(int _n) : n(_n) { t.resize(n); }
  fenwick(const vector<C>& a) : n(a.size()) {
    t.resize(n);
    for (int i = 0; i < n; ++i) {
      upd(i, a[i]);
    }
  }
  C get(int k) {
    C res = 0;
    for (k; k >= 0; k = (k & (k + 1)) - 1) {
      res += t[k];
    }
    return res;
  }
  void upd(int k, C x) {
    for (k; k < n; k |= k + 1) {
      t[k] += x;
    }
  }
  C get(int l, int r) { return get(r) - get(l - 1); }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> x(n), v(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
  }
  map<int, int> mapchik;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    mapchik[v[i]] = 1;
  }
  int sz = 0;
  for (auto& [a, b] : mapchik) {
    b = sz++;
  }
  for (int i = 0; i < n; ++i) {
    v[i] = mapchik[v[i]];
  }
  vector<int> ord(n);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(), [&](int i, int j) { return x[i] < x[j]; });
  long long ans = 0;
  fenwick<int> cnt(sz);
  fenwick<long long> sum(sz);
  for (int i : ord) {
    ans += 1ll * cnt.get(0, v[i]) * x[i] - sum.get(0, v[i]);
    cnt.upd(v[i], 1);
    sum.upd(v[i], x[i]);
  }
  cout << ans;
  return 0;
}

```