#include <bits/stdc++.h>
using namespace std;
long long n, ans, tree[800002], tree1[800002];
map<long long, long long> mp, mp1;
pair<long long, long long> a[300005];
vector<long long> v;
void inc1(int idx, int val) {
  for (int i = idx; i <= n; i += i & (-i)) {
    tree[i] += val;
  }
}
void inc2(int idx, int val) {
  for (int i = idx; i <= n; i += (i & (-i))) {
    tree1[i] += val;
  }
}
long long sum1(int idx) {
  long long pas = 0;
  for (int i = idx; i > 0; i -= i & (-i)) {
    pas += tree[i];
  }
  return pas;
}
long long sum2(int idx) {
  long long pas = 0;
  for (int i = idx; i > 0; i -= i & (-i)) {
    pas += tree1[i];
  }
  return pas;
}
int main() {
  std::ios_base::sync_with_stdio(false), cin.tie(0), cout.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;
    if (!mp[a[i].second]) {
      v.push_back(a[i].second);
    }
    mp[a[i].second] = 1;
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    mp1[v[i]] = i + 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    ans += sum1(mp1[a[i].second]) * a[i].first - sum2(mp1[a[i].second]);
    inc1(mp1[a[i].second], 1);
    inc2(mp1[a[i].second], a[i].first);
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long n, ans, tree[800002], tree1[800002];
map<long long, long long> mp, mp1;
pair<long long, long long> a[300005];
vector<long long> v;
void inc1(int idx, int val) {
  for (int i = idx; i <= n; i += i & (-i)) {
    tree[i] += val;
  }
}
void inc2(int idx, int val) {
  for (int i = idx; i <= n; i += (i & (-i))) {
    tree1[i] += val;
  }
}
long long sum1(int idx) {
  long long pas = 0;
  for (int i = idx; i > 0; i -= i & (-i)) {
    pas += tree[i];
  }
  return pas;
}
long long sum2(int idx) {
  long long pas = 0;
  for (int i = idx; i > 0; i -= i & (-i)) {
    pas += tree1[i];
  }
  return pas;
}
int main() {
  std::ios_base::sync_with_stdio(false), cin.tie(0), cout.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;
    if (!mp[a[i].second]) {
      v.push_back(a[i].second);
    }
    mp[a[i].second] = 1;
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    mp1[v[i]] = i + 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    ans += sum1(mp1[a[i].second]) * a[i].first - sum2(mp1[a[i].second]);
    inc1(mp1[a[i].second], 1);
    inc2(mp1[a[i].second], a[i].first);
  }
  cout << ans << endl;
}

```