#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 5;
int n, t[M], k;
long long ans;
struct node {
  long long p;
  int v;
} a[M];
struct BIT {
  long long s[M];
  void add(int i, long long x) {
    for (; i <= k; i += i & -i) s[i] += x;
  }
  long long ask(int i) {
    long long res = 0;
    for (; i > 0; i -= i & -i) res += s[i];
    return res;
  }
} f1, f2;
inline int find(int x) {
  int l = 1, r = k, mid;
  while (l < r) {
    mid = (l + r) >> 1;
    if (t[mid] < x)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
int cmp(node x, node y) { return x.p < y.p; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].p);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].v);
    t[i] = a[i].v;
  }
  sort(a + 1, a + n + 1, cmp);
  sort(t + 1, t + n + 1);
  k++;
  for (int i = 2; i <= n; i++)
    if (t[i] != t[i - 1]) t[++k] = t[i];
  for (int i = 1; i <= n; i++) {
    int w = find(a[i].v);
    ans += a[i].p * f1.ask(w) - f2.ask(w);
    f1.add(w, 1);
    f2.add(w, a[i].p);
  }
  cout << ans << endl;
  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 M = 2e5 + 5;
int n, t[M], k;
long long ans;
struct node {
  long long p;
  int v;
} a[M];
struct BIT {
  long long s[M];
  void add(int i, long long x) {
    for (; i <= k; i += i & -i) s[i] += x;
  }
  long long ask(int i) {
    long long res = 0;
    for (; i > 0; i -= i & -i) res += s[i];
    return res;
  }
} f1, f2;
inline int find(int x) {
  int l = 1, r = k, mid;
  while (l < r) {
    mid = (l + r) >> 1;
    if (t[mid] < x)
      l = mid + 1;
    else
      r = mid;
  }
  return l;
}
int cmp(node x, node y) { return x.p < y.p; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i].p);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].v);
    t[i] = a[i].v;
  }
  sort(a + 1, a + n + 1, cmp);
  sort(t + 1, t + n + 1);
  k++;
  for (int i = 2; i <= n; i++)
    if (t[i] != t[i - 1]) t[++k] = t[i];
  for (int i = 1; i <= n; i++) {
    int w = find(a[i].v);
    ans += a[i].p * f1.ask(w) - f2.ask(w);
    f1.add(w, 1);
    f2.add(w, a[i].p);
  }
  cout << ans << endl;
  return 0;
}

```