#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 1e6 + 5;
const long long M = 1e8;
const long long mod = 1e9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, a[N], v[N], ct = 1;
vector<long long> sp;
vector<pair<long long, long long> > x;
map<long long, long long> ind;
long long tree[N][2] = {};
void update(int idx, int val, int first) {
  idx++;
  while (idx <= n + 1) {
    tree[idx][first] += val;
    idx += idx & (-idx);
  }
}
long long query(int idx, int first) {
  idx++;
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx][first];
    idx -= idx & (-idx);
  }
  return sum;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> v[i], sp.push_back(v[i]);
  sort(sp.begin(), sp.end());
  for (auto it : sp)
    if (ind[it] == 0) ind[it] = ct++;
  for (int i = 1; i <= n; i++) v[i] = ind[v[i]], x.push_back({a[i], v[i]});
  sort(x.begin(), x.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += query(x[i].second, 0) * x[i].first - query(x[i].second, 1);
    update(x[i].second, 1, 0);
    update(x[i].second, x[i].first, 1);
  }
  cout << ans << " "
       << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

### 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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 1e6 + 5;
const long long M = 1e8;
const long long mod = 1e9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, a[N], v[N], ct = 1;
vector<long long> sp;
vector<pair<long long, long long> > x;
map<long long, long long> ind;
long long tree[N][2] = {};
void update(int idx, int val, int first) {
  idx++;
  while (idx <= n + 1) {
    tree[idx][first] += val;
    idx += idx & (-idx);
  }
}
long long query(int idx, int first) {
  idx++;
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx][first];
    idx -= idx & (-idx);
  }
  return sum;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> v[i], sp.push_back(v[i]);
  sort(sp.begin(), sp.end());
  for (auto it : sp)
    if (ind[it] == 0) ind[it] = ct++;
  for (int i = 1; i <= n; i++) v[i] = ind[v[i]], x.push_back({a[i], v[i]});
  sort(x.begin(), x.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += query(x[i].second, 0) * x[i].first - query(x[i].second, 1);
    update(x[i].second, 1, 0);
    update(x[i].second, x[i].first, 1);
  }
  cout << ans << " "
       << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

```