#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[200005];
long long dp[200005];
long long tr[2][200005];
void update(long long f, long long x, long long val) {
  while (x < 200005) {
    tr[f][x] += val;
    x += x & (-x);
  }
}
long long query(long long f, long long x) {
  long long s = 0;
  while (x > 0) {
    s += tr[f][x];
    x -= x & (-x);
  }
  return s;
}
int main() {
  long long n, i, j;
  string t;
  while (cin >> n) {
    for (i = 1; i <= n; i++) cin >> a[i];
    vector<pair<long long, long long> > v;
    map<long long, long long> mp;
    for (i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      v.push_back(make_pair(a[i], x));
      mp[x];
    }
    map<long long, long long>::iterator it;
    sort(v.begin(), v.end());
    long long cn = 0;
    for (it = mp.begin(); it != mp.end(); it++) it->second = ++cn;
    long long ans = 0;
    reverse(v.begin(), v.end());
    for (i = 0; i < v.size(); i++) {
      long long p = v[i].first;
      long long q = mp[v[i].second];
      long long r = query(0, cn) - query(0, q - 1);
      long long r1 = query(1, cn) - query(1, q - 1);
      update(0, q, 1);
      update(1, q, p);
      ans += (r1 - p * r);
    }
    cout << ans << endl;
  }
  return 0;
}

### 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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[200005];
long long dp[200005];
long long tr[2][200005];
void update(long long f, long long x, long long val) {
  while (x < 200005) {
    tr[f][x] += val;
    x += x & (-x);
  }
}
long long query(long long f, long long x) {
  long long s = 0;
  while (x > 0) {
    s += tr[f][x];
    x -= x & (-x);
  }
  return s;
}
int main() {
  long long n, i, j;
  string t;
  while (cin >> n) {
    for (i = 1; i <= n; i++) cin >> a[i];
    vector<pair<long long, long long> > v;
    map<long long, long long> mp;
    for (i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      v.push_back(make_pair(a[i], x));
      mp[x];
    }
    map<long long, long long>::iterator it;
    sort(v.begin(), v.end());
    long long cn = 0;
    for (it = mp.begin(); it != mp.end(); it++) it->second = ++cn;
    long long ans = 0;
    reverse(v.begin(), v.end());
    for (i = 0; i < v.size(); i++) {
      long long p = v[i].first;
      long long q = mp[v[i].second];
      long long r = query(0, cn) - query(0, q - 1);
      long long r1 = query(1, cn) - query(1, q - 1);
      update(0, q, 1);
      update(1, q, p);
      ans += (r1 - p * r);
    }
    cout << ans << endl;
  }
  return 0;
}

```