#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
long long fwk[maxn], cnt[maxn];
void upd(int p, int c) {
  for (; p < maxn; p += p & -p) fwk[p] += c, cnt[p]++;
}
void qry(int p, long long& sum, long long& c) {
  sum = c = 0;
  for (; p; p -= p & -p) sum += fwk[p], c += cnt[p];
}
vector<int> d;
inline int hs(int x) {
  return lower_bound(d.begin(), d.end(), x) - d.begin() + 1;
}
pair<int, int> p[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i].first);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i].second);
    d.push_back(p[i].second);
  }
  sort(p + 1, p + 1 + n);
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  long long ans = 0, sum, c;
  for (int i = 1; i <= n; ++i) {
    qry(hs(p[i].second), sum, c);
    ans += c * p[i].first - sum;
    upd(hs(p[i].second), p[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 maxn = 2e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
long long fwk[maxn], cnt[maxn];
void upd(int p, int c) {
  for (; p < maxn; p += p & -p) fwk[p] += c, cnt[p]++;
}
void qry(int p, long long& sum, long long& c) {
  sum = c = 0;
  for (; p; p -= p & -p) sum += fwk[p], c += cnt[p];
}
vector<int> d;
inline int hs(int x) {
  return lower_bound(d.begin(), d.end(), x) - d.begin() + 1;
}
pair<int, int> p[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &p[i].first);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i].second);
    d.push_back(p[i].second);
  }
  sort(p + 1, p + 1 + n);
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  long long ans = 0, sum, c;
  for (int i = 1; i <= n; ++i) {
    qry(hs(p[i].second), sum, c);
    ans += c * p[i].first - sum;
    upd(hs(p[i].second), p[i].first);
  }
  cout << ans;
  return 0;
}

```