#include <bits/stdc++.h>
using namespace std;
template <typename T>
using pr = pair<T, T>;
const int mod = 1e9 + 7;
struct SegTree {
  vector<pr<long long>> segtree;
  long long N;
  SegTree(long long n) {
    N = 1;
    while (N < n) N <<= 1;
    segtree = vector<pr<long long>>(2 * N);
  }
  void modify(long long k, long long x) {
    k += N;
    segtree[k].second += x;
    segtree[k].first += 1;
    for (k /= 2; k > 0; k /= 2) {
      segtree[k].first = segtree[k * 2].first + segtree[k * 2 + 1].first;
      segtree[k].second = segtree[k * 2].second + segtree[k * 2 + 1].second;
    }
  }
  pr<long long> query(long long b) {
    long long a = 0;
    long long ansx = 0, ansy = 0;
    for (a += N, b += N; a <= b; a /= 2, b /= 2) {
      if (a % 2 == 1) ansx += segtree[a].first, ansy += segtree[a++].second;
      if (b % 2 == 0) ansx += segtree[b].first, ansy += segtree[b--].second;
    }
    return make_pair(ansx, ansy);
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<pr<long long>> xv(n);
  vector<long long> v(n);
  for (signed i = 0; i < n; ++i) cin >> xv[i].first;
  for (signed i = 0; i < n; ++i) cin >> xv[i].second, v[i] = xv[i].second;
  sort((xv).begin(), (xv).end());
  sort((v).begin(), (v).end());
  map<long long, int> cmp;
  for (signed i = 0; i < n; ++i) cmp[v[i]] = i;
  long long ans = 0;
  SegTree x(n);
  x.modify(cmp[xv[0].second], xv[0].first);
  for (signed i = 1; i < n; ++i) {
    pr<long long> qa = x.query(cmp[xv[i].second]);
    ans += qa.first * xv[i].first - qa.second;
    x.modify(cmp[xv[i].second], xv[i].first);
  }
  cout << ans << endl;
}
signed main() {
  signed t = 1;
  while (t--) solve();
}

### Prompt

Generate 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;
template <typename T>
using pr = pair<T, T>;
const int mod = 1e9 + 7;
struct SegTree {
  vector<pr<long long>> segtree;
  long long N;
  SegTree(long long n) {
    N = 1;
    while (N < n) N <<= 1;
    segtree = vector<pr<long long>>(2 * N);
  }
  void modify(long long k, long long x) {
    k += N;
    segtree[k].second += x;
    segtree[k].first += 1;
    for (k /= 2; k > 0; k /= 2) {
      segtree[k].first = segtree[k * 2].first + segtree[k * 2 + 1].first;
      segtree[k].second = segtree[k * 2].second + segtree[k * 2 + 1].second;
    }
  }
  pr<long long> query(long long b) {
    long long a = 0;
    long long ansx = 0, ansy = 0;
    for (a += N, b += N; a <= b; a /= 2, b /= 2) {
      if (a % 2 == 1) ansx += segtree[a].first, ansy += segtree[a++].second;
      if (b % 2 == 0) ansx += segtree[b].first, ansy += segtree[b--].second;
    }
    return make_pair(ansx, ansy);
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<pr<long long>> xv(n);
  vector<long long> v(n);
  for (signed i = 0; i < n; ++i) cin >> xv[i].first;
  for (signed i = 0; i < n; ++i) cin >> xv[i].second, v[i] = xv[i].second;
  sort((xv).begin(), (xv).end());
  sort((v).begin(), (v).end());
  map<long long, int> cmp;
  for (signed i = 0; i < n; ++i) cmp[v[i]] = i;
  long long ans = 0;
  SegTree x(n);
  x.modify(cmp[xv[0].second], xv[0].first);
  for (signed i = 1; i < n; ++i) {
    pr<long long> qa = x.query(cmp[xv[i].second]);
    ans += qa.first * xv[i].first - qa.second;
    x.modify(cmp[xv[i].second], xv[i].first);
  }
  cout << ans << endl;
}
signed main() {
  signed t = 1;
  while (t--) solve();
}

```