#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)2e5 + 1;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
pair<pair<long long, long long>, int> xv[MAXN];
long long n, BIT[MAXN], BIT2[MAXN];
void update(int x, long long val) {
  for (; x <= n; x += x & -x) {
    BIT[x] += val;
    BIT2[x]++;
  }
}
pair<long long, long long> query(int x) {
  long long sum = 0;
  long long num = 0;
  for (; x > 0; x -= x & -x) {
    sum += BIT[x];
    num += BIT2[x];
  }
  return {sum, num};
}
bool comp(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  if (a.first.second == b.first.second) return a.first.first < b.first.first;
  return a.first.second < b.first.second;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> xv[i].first.first;
  }
  for (int i = 0; i < n; i++) {
    cin >> xv[i].first.second;
  }
  sort(xv, xv + n);
  for (int i = 0; i < n; i++) {
    xv[i].second = i + 1;
  }
  long long ans = 0;
  sort(xv, xv + n, comp);
  for (int i = 0; i < n; i++) {
    pair<long long, long long> res = query(xv[i].second);
    ans += (res.second * xv[i].first.first - res.first);
    update(xv[i].second, xv[i].first.first);
  }
  cout << ans;
  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>
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)2e5 + 1;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
pair<pair<long long, long long>, int> xv[MAXN];
long long n, BIT[MAXN], BIT2[MAXN];
void update(int x, long long val) {
  for (; x <= n; x += x & -x) {
    BIT[x] += val;
    BIT2[x]++;
  }
}
pair<long long, long long> query(int x) {
  long long sum = 0;
  long long num = 0;
  for (; x > 0; x -= x & -x) {
    sum += BIT[x];
    num += BIT2[x];
  }
  return {sum, num};
}
bool comp(pair<pair<long long, long long>, int> a,
          pair<pair<long long, long long>, int> b) {
  if (a.first.second == b.first.second) return a.first.first < b.first.first;
  return a.first.second < b.first.second;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> xv[i].first.first;
  }
  for (int i = 0; i < n; i++) {
    cin >> xv[i].first.second;
  }
  sort(xv, xv + n);
  for (int i = 0; i < n; i++) {
    xv[i].second = i + 1;
  }
  long long ans = 0;
  sort(xv, xv + n, comp);
  for (int i = 0; i < n; i++) {
    pair<long long, long long> res = query(xv[i].second);
    ans += (res.second * xv[i].first.first - res.first);
    update(xv[i].second, xv[i].first.first);
  }
  cout << ans;
  return 0;
}

```