#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const long long mod = 998244353;
int T, n;
struct node {
  long long d, v;
  friend bool operator<(node a, node b) { return a.d < b.d; }
} a[maxn];
long long c[maxn], d[maxn];
int b[maxn];
void update(int x, int y) {
  for (int i = x; i <= n; i += (i & (-i))) c[i] += y, d[i]++;
}
long long query1(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & (-i))) ans += c[i];
  return ans;
}
long long query2(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & (-i))) ans += d[i];
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].d);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].v), b[i] = a[i].v;
  sort(b + 1, b + 1 + n);
  int m = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i) {
    a[i].v = lower_bound(b + 1, b + 1 + m, a[i].v) - b;
  }
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += (query2(a[i].v) * a[i].d - query1(a[i].v));
    update(a[i].v, a[i].d);
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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 maxn = 2e5 + 5;
const long long mod = 998244353;
int T, n;
struct node {
  long long d, v;
  friend bool operator<(node a, node b) { return a.d < b.d; }
} a[maxn];
long long c[maxn], d[maxn];
int b[maxn];
void update(int x, int y) {
  for (int i = x; i <= n; i += (i & (-i))) c[i] += y, d[i]++;
}
long long query1(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & (-i))) ans += c[i];
  return ans;
}
long long query2(int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= (i & (-i))) ans += d[i];
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].d);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].v), b[i] = a[i].v;
  sort(b + 1, b + 1 + n);
  int m = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i) {
    a[i].v = lower_bound(b + 1, b + 1 + m, a[i].v) - b;
  }
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += (query2(a[i].v) * a[i].d - query1(a[i].v));
    update(a[i].v, a[i].d);
  }
  cout << ans;
  return 0;
}

```