#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int n;
long long ans = 0;
vector<long long> T1(N, 0), T2(N, 0);
pair<int, int> a[N];
vector<int> b;
void update(vector<long long>& T, int x, long long v) {
  for (; x <= (int)b.size(); x += (x & (-x))) T[x] += v;
}
long long get(const vector<long long>& T, int x) {
  long long res = 0;
  for (; x; x -= (x & (-x))) res += T[x];
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> a[i].first;
  for (int i = (0); i < (n); i++) {
    cin >> a[i].second;
    b.push_back(a[i].second);
  }
  sort(a, a + n);
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  int vt = lower_bound(b.begin(), b.end(), a[0].second) - b.begin() + 1;
  update(T1, vt, 1ll * a[0].first);
  update(T2, vt, 1);
  for (int i = (1); i < (n); i++) {
    vt = lower_bound(b.begin(), b.end(), a[i].second) - b.begin() + 1;
    ans += get(T2, vt) * a[i].first - get(T1, vt);
    update(T1, vt, 1ll * a[i].first);
    update(T2, vt, 1);
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 2e5 + 7;
int n;
long long ans = 0;
vector<long long> T1(N, 0), T2(N, 0);
pair<int, int> a[N];
vector<int> b;
void update(vector<long long>& T, int x, long long v) {
  for (; x <= (int)b.size(); x += (x & (-x))) T[x] += v;
}
long long get(const vector<long long>& T, int x) {
  long long res = 0;
  for (; x; x -= (x & (-x))) res += T[x];
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> a[i].first;
  for (int i = (0); i < (n); i++) {
    cin >> a[i].second;
    b.push_back(a[i].second);
  }
  sort(a, a + n);
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  int vt = lower_bound(b.begin(), b.end(), a[0].second) - b.begin() + 1;
  update(T1, vt, 1ll * a[0].first);
  update(T2, vt, 1);
  for (int i = (1); i < (n); i++) {
    vt = lower_bound(b.begin(), b.end(), a[i].second) - b.begin() + 1;
    ans += get(T2, vt) * a[i].first - get(T1, vt);
    update(T1, vt, 1ll * a[i].first);
    update(T2, vt, 1);
  }
  cout << ans;
  return 0;
}

```