#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> a[200005];
long long sum[1000005], cnt[1000005];
long long querysum(long long pos) {
  long long ret = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    ret += sum[pos];
  }
  return ret;
}
long long querycnt(long long pos) {
  long long ret = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    ret += cnt[pos];
  }
  return ret;
}
void updatesum(long long pos, long long val) {
  for (; pos < 1000000; pos |= pos + 1) {
    sum[pos] += val;
  }
}
void updatecnt(long long pos, long long val) {
  for (; pos < 1000000; pos |= pos + 1) {
    cnt[pos] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  vector<long long> v;
  for (long long i = 1; i <= n; i++) {
    v.push_back(a[i].second);
  }
  sort(v.begin(), v.end());
  map<long long, long long> num;
  for (long long i = 0; i < v.size(); i++) {
    num[v[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    a[i].second = num[a[i].second];
  }
  sort(a + 1, a + n + 1);
  memset(sum, 0, sizeof(sum));
  memset(cnt, 0, sizeof(cnt));
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += a[i].first * querycnt(a[i].second) - querysum(a[i].second);
    updatesum(a[i].second, a[i].first);
    updatecnt(a[i].second, 1);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
pair<long long, long long> a[200005];
long long sum[1000005], cnt[1000005];
long long querysum(long long pos) {
  long long ret = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    ret += sum[pos];
  }
  return ret;
}
long long querycnt(long long pos) {
  long long ret = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    ret += cnt[pos];
  }
  return ret;
}
void updatesum(long long pos, long long val) {
  for (; pos < 1000000; pos |= pos + 1) {
    sum[pos] += val;
  }
}
void updatecnt(long long pos, long long val) {
  for (; pos < 1000000; pos |= pos + 1) {
    cnt[pos] += val;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i].first;
  for (long long i = 1; i <= n; i++) cin >> a[i].second;
  vector<long long> v;
  for (long long i = 1; i <= n; i++) {
    v.push_back(a[i].second);
  }
  sort(v.begin(), v.end());
  map<long long, long long> num;
  for (long long i = 0; i < v.size(); i++) {
    num[v[i]] = i;
  }
  for (long long i = 1; i <= n; i++) {
    a[i].second = num[a[i].second];
  }
  sort(a + 1, a + n + 1);
  memset(sum, 0, sizeof(sum));
  memset(cnt, 0, sizeof(cnt));
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += a[i].first * querycnt(a[i].second) - querysum(a[i].second);
    updatesum(a[i].second, a[i].first);
    updatecnt(a[i].second, 1);
  }
  cout << ans << endl;
  return 0;
}

```