#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long>> v;
long long arr[200005];
map<long long, long long> mp;
map<long long, long long> rmp;
pair<long long, long long> bit[200005];
set<long long> s;
void insert(long long ind) {
  for (long long i = ind; i < 200005; i += i & -i)
    bit[i].first += 1, bit[i].second += rmp[ind];
}
pair<long long, long long> query(long long ind) {
  pair<long long, long long> ret = {0, 0};
  if (ind == 0) return ret;
  for (long long i = ind; i > 0; i -= i & -i)
    ret.first += bit[i].first, ret.second += bit[i].second;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> arr[i], s.insert(arr[i]);
  long long c = 1;
  for (auto i : s) rmp[c] = i, mp[i] = c++;
  for (long long i = 0; i < n; i++) {
    long long ve;
    cin >> ve;
    v.push_back({ve, mp[arr[i]]});
  }
  sort(v.begin(), v.end(), greater<pair<long long, long long>>());
  long long ans = 0;
  c = 0;
  long long val = 0;
  for (auto i : v) {
    pair<long long, long long> q = query(i.second);
    long long num = c - q.first;
    long long sum = val - q.second;
    ans += (sum - num * rmp[i.second]);
    c++;
    val += rmp[i.second];
    insert(i.second);
  }
  cout << ans << "\n";
}

### 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;
vector<pair<long long, long long>> v;
long long arr[200005];
map<long long, long long> mp;
map<long long, long long> rmp;
pair<long long, long long> bit[200005];
set<long long> s;
void insert(long long ind) {
  for (long long i = ind; i < 200005; i += i & -i)
    bit[i].first += 1, bit[i].second += rmp[ind];
}
pair<long long, long long> query(long long ind) {
  pair<long long, long long> ret = {0, 0};
  if (ind == 0) return ret;
  for (long long i = ind; i > 0; i -= i & -i)
    ret.first += bit[i].first, ret.second += bit[i].second;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> arr[i], s.insert(arr[i]);
  long long c = 1;
  for (auto i : s) rmp[c] = i, mp[i] = c++;
  for (long long i = 0; i < n; i++) {
    long long ve;
    cin >> ve;
    v.push_back({ve, mp[arr[i]]});
  }
  sort(v.begin(), v.end(), greater<pair<long long, long long>>());
  long long ans = 0;
  c = 0;
  long long val = 0;
  for (auto i : v) {
    pair<long long, long long> q = query(i.second);
    long long num = c - q.first;
    long long sum = val - q.second;
    ans += (sum - num * rmp[i.second]);
    c++;
    val += rmp[i.second];
    insert(i.second);
  }
  cout << ans << "\n";
}

```