#include <bits/stdc++.h>
using namespace std;
long long n;
pair<long long, long long> p[200005];
long long v[200005];
map<long long, long long> m;
long long f0[200005];
long long f1[200005];
void up0(long long x, long long val) {
  for (long long i = x; i <= 200000; i += i & -i) f0[i] += val;
}
void up1(long long x, long long val) {
  for (long long i = x; i <= 200000; i += i & -i) f1[i] += val;
}
long long get0(long long pos) {
  long long ans = 0;
  for (long long i = pos; i > 0; i -= i & -i) ans += f0[i];
  return ans;
}
long long get1(long long pos) {
  long long ans = 0;
  for (long long i = pos; i > 0; i -= i & -i) ans += f1[i];
  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 >> p[i].first;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].second;
    v[i] = p[i].second;
  }
  sort(v + 1, v + 1 + n);
  sort(p + 1, p + 1 + n);
  long long c = 0;
  for (long long i = 1; i <= n; i++) {
    if (m[v[i]] == 0) {
      m[v[i]] = ++c;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long x = p[i].first;
    long long sp = p[i].second;
    long long c = get0(m[sp]);
    long long cur = get1(m[sp]);
    ans += x * c - cur;
    up0(m[sp], 1);
    up1(m[sp], x);
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long n;
pair<long long, long long> p[200005];
long long v[200005];
map<long long, long long> m;
long long f0[200005];
long long f1[200005];
void up0(long long x, long long val) {
  for (long long i = x; i <= 200000; i += i & -i) f0[i] += val;
}
void up1(long long x, long long val) {
  for (long long i = x; i <= 200000; i += i & -i) f1[i] += val;
}
long long get0(long long pos) {
  long long ans = 0;
  for (long long i = pos; i > 0; i -= i & -i) ans += f0[i];
  return ans;
}
long long get1(long long pos) {
  long long ans = 0;
  for (long long i = pos; i > 0; i -= i & -i) ans += f1[i];
  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 >> p[i].first;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i].second;
    v[i] = p[i].second;
  }
  sort(v + 1, v + 1 + n);
  sort(p + 1, p + 1 + n);
  long long c = 0;
  for (long long i = 1; i <= n; i++) {
    if (m[v[i]] == 0) {
      m[v[i]] = ++c;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long x = p[i].first;
    long long sp = p[i].second;
    long long c = get0(m[sp]);
    long long cur = get1(m[sp]);
    ans += x * c - cur;
    up0(m[sp], 1);
    up1(m[sp], x);
  }
  cout << ans;
  return 0;
}

```