#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > v;
vector<long long> v1;
long long a[200005], b[200005], c[200005], d[200005], n, m;
map<long long, long long> mp;
long long get(long long f[], long long pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
void upd(long long f[], long long pos, long long val) {
  for (; pos < m; pos |= pos + 1) f[pos] += val;
}
int main() {
  long long i;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) cin >> b[i];
  for (i = 0; i < n; i++) {
    v.push_back({a[i], b[i]});
    if (!mp[b[i]]) v1.push_back(b[i]);
    mp[b[i]] = 1;
  }
  sort(v.begin(), v.end());
  sort(v1.begin(), v1.end());
  m = v1.size();
  long long s = 0;
  for (auto x : v) {
    long long y = lower_bound(v1.begin(), v1.end(), x.second) - v1.begin();
    s += get(c, y) * x.first - get(d, y);
    upd(c, y, 1);
    upd(d, y, x.first);
  }
  cout << s;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<pair<long long, long long> > v;
vector<long long> v1;
long long a[200005], b[200005], c[200005], d[200005], n, m;
map<long long, long long> mp;
long long get(long long f[], long long pos) {
  long long res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
void upd(long long f[], long long pos, long long val) {
  for (; pos < m; pos |= pos + 1) f[pos] += val;
}
int main() {
  long long i;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) cin >> b[i];
  for (i = 0; i < n; i++) {
    v.push_back({a[i], b[i]});
    if (!mp[b[i]]) v1.push_back(b[i]);
    mp[b[i]] = 1;
  }
  sort(v.begin(), v.end());
  sort(v1.begin(), v1.end());
  m = v1.size();
  long long s = 0;
  for (auto x : v) {
    long long y = lower_bound(v1.begin(), v1.end(), x.second) - v1.begin();
    s += get(c, y) * x.first - get(d, y);
    upd(c, y, 1);
    upd(d, y, x.first);
  }
  cout << s;
  return 0;
}

```