#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct FTDown {
  long long ft[N + 5];
  FTDown() {
    for (int i = 1; i <= N; i++) ft[i] = 0;
  }
  void update(int p, long long v) {
    for (; p <= N; p += (p & -p)) ft[p] += v;
  }
  long long get(int p) {
    long long sum = 0;
    for (; p; p -= (p & -p)) sum += ft[p];
    return sum;
  }
};
int n, b[N];
pair<int, int> a[N];
FTDown sumXDown, cntDown;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
    b[i] = a[i].second;
  }
  b[n + 1] = 0;
  sort(b + 1, b + n + 2);
  int m = 1;
  for (int i = 2; i <= n + 1; i++)
    if (b[i] != b[m]) b[++m] = b[i];
  for (int i = 1; i <= n; i++)
    a[i].second = lower_bound(b + 1, b + m + 1, a[i].second) - b;
  sort(a + 1, a + n + 1);
  int ind0 = lower_bound(b + 1, b + m + 1, 0) - b;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long cnt = cntDown.get(a[i].second);
    long long sumX = sumXDown.get(a[i].second);
    ans += cnt * a[i].first - sumX;
    cntDown.update(a[i].second, 1);
    sumXDown.update(a[i].second, a[i].first);
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in CPP for 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 + 5;
struct FTDown {
  long long ft[N + 5];
  FTDown() {
    for (int i = 1; i <= N; i++) ft[i] = 0;
  }
  void update(int p, long long v) {
    for (; p <= N; p += (p & -p)) ft[p] += v;
  }
  long long get(int p) {
    long long sum = 0;
    for (; p; p -= (p & -p)) sum += ft[p];
    return sum;
  }
};
int n, b[N];
pair<int, int> a[N];
FTDown sumXDown, cntDown;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].first;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
    b[i] = a[i].second;
  }
  b[n + 1] = 0;
  sort(b + 1, b + n + 2);
  int m = 1;
  for (int i = 2; i <= n + 1; i++)
    if (b[i] != b[m]) b[++m] = b[i];
  for (int i = 1; i <= n; i++)
    a[i].second = lower_bound(b + 1, b + m + 1, a[i].second) - b;
  sort(a + 1, a + n + 1);
  int ind0 = lower_bound(b + 1, b + m + 1, 0) - b;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long cnt = cntDown.get(a[i].second);
    long long sumX = sumXDown.get(a[i].second);
    ans += cnt * a[i].first - sumX;
    cntDown.update(a[i].second, 1);
    sumXDown.update(a[i].second, a[i].first);
  }
  cout << ans;
  return 0;
}

```