#include <bits/stdc++.h>
using namespace std;
long long n;
pair<long long, long long> a[200001];
long long cnt[200001], sum[200001];
void update(long long x, long long val) {
  for (; x <= 200000; x += x & (-x)) {
    cnt[x]++;
    sum[x] += val;
  }
}
long long query(long long x, long long val) {
  long long ans = 0;
  for (; x > 0; x -= x & (-x)) ans += (cnt[x] * val - sum[x]);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  vector<pair<long long, long long> > val;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + 1 + n);
  for (long long i = 1; i <= n; i++)
    val.push_back(pair<long long, long long>(a[i].second, i));
  sort(val.begin(), val.end());
  long long pre = -1e9, top = 0;
  for (long long i = 0; i < val.size(); i++) {
    if (val[i].first != pre) {
      pre = val[i].first;
      top++;
    }
    a[val[i].second].second = top;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += query(a[i].second, a[i].first);
    update(a[i].second, a[i].first);
  }
  cout << ans;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
pair<long long, long long> a[200001];
long long cnt[200001], sum[200001];
void update(long long x, long long val) {
  for (; x <= 200000; x += x & (-x)) {
    cnt[x]++;
    sum[x] += val;
  }
}
long long query(long long x, long long val) {
  long long ans = 0;
  for (; x > 0; x -= x & (-x)) ans += (cnt[x] * val - sum[x]);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  vector<pair<long long, long long> > val;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  sort(a + 1, a + 1 + n);
  for (long long i = 1; i <= n; i++)
    val.push_back(pair<long long, long long>(a[i].second, i));
  sort(val.begin(), val.end());
  long long pre = -1e9, top = 0;
  for (long long i = 0; i < val.size(); i++) {
    if (val[i].first != pre) {
      pre = val[i].first;
      top++;
    }
    a[val[i].second].second = top;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += query(a[i].second, a[i].first);
    update(a[i].second, a[i].first);
  }
  cout << ans;
}

```