#include <bits/stdc++.h>
using namespace std;
int n, op;
pair<long long, int> tree[201000];
pair<long long, int> q[201000];
long long a[201000];
int lowbits(int x) { return x & (-x); }
void add(long long x, int pos) {
  while (pos <= n) {
    tree[pos].first += x;
    tree[pos].second++;
    pos += lowbits(pos);
  }
}
pair<long long, int> getsum(int x) {
  pair<long long, int> res;
  while (x > 0) {
    res.first += tree[x].first;
    res.second += tree[x].second;
    x -= lowbits(x);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &q[i].first);
  for (int i = 1; i <= n; i++) scanf("%d", &q[i].second), a[++op] = q[i].second;
  sort(a + 1, a + 1 + n), sort(q + 1, q + 1 + n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(a + 1, a + 1 + n, q[i].second) - a;
    pair<long long, int> now = getsum(pos);
    ans += now.second * q[i].first - now.first;
    add(q[i].first, pos);
  }
  cout << ans << endl;
}

### 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;
int n, op;
pair<long long, int> tree[201000];
pair<long long, int> q[201000];
long long a[201000];
int lowbits(int x) { return x & (-x); }
void add(long long x, int pos) {
  while (pos <= n) {
    tree[pos].first += x;
    tree[pos].second++;
    pos += lowbits(pos);
  }
}
pair<long long, int> getsum(int x) {
  pair<long long, int> res;
  while (x > 0) {
    res.first += tree[x].first;
    res.second += tree[x].second;
    x -= lowbits(x);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &q[i].first);
  for (int i = 1; i <= n; i++) scanf("%d", &q[i].second), a[++op] = q[i].second;
  sort(a + 1, a + 1 + n), sort(q + 1, q + 1 + n);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(a + 1, a + 1 + n, q[i].second) - a;
    pair<long long, int> now = getsum(pos);
    ans += now.second * q[i].first - now.first;
    add(q[i].first, pos);
  }
  cout << ans << endl;
}

```