#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n;
long long ans;
pair<long long, long long> a[N];
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  if (x.second != y.second) {
    return x.second < y.second;
  } else {
    return x.first < y.first;
  }
}
vector<long long> segmenttree[4 * N], segmenttree2[4 * N];
void build(int id, int l, int r) {
  if (l == r) {
    segmenttree[id] = {a[l].first};
    segmenttree2[id] = {a[l].first};
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  merge(segmenttree[id << 1].begin(), segmenttree[id << 1].end(),
        segmenttree[id << 1 | 1].begin(), segmenttree[id << 1 | 1].end(),
        back_inserter(segmenttree[id]));
  segmenttree2[id].resize(r - l + 1);
  segmenttree2[id][r - l] = segmenttree[id][r - l];
  for (int i = r - l - 1; i >= 0; i--) {
    segmenttree2[id][i] = segmenttree2[id][i + 1] + segmenttree[id][i];
  }
}
long long get(int id, int l, int r, int u, int v, long long val) {
  if (v < l || r < u) {
    return 0;
  }
  if (u <= l && r <= v) {
    long long idx =
        lower_bound(segmenttree[id].begin(), segmenttree[id].end(), val) -
        segmenttree[id].begin();
    if (idx == r - l + 1) {
      return 0;
    } else {
      return segmenttree2[id][idx] - (r - l + 1 - idx) * val;
    }
  }
  int mid = (l + r) >> 1;
  return get(id << 1, l, mid, u, v, val) +
         get(id << 1 | 1, mid + 1, r, u, v, val);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
  }
  sort(a + 1, a + 1 + n, cmp);
  build(1, 1, n);
  for (int i = 1; i < n; i++) {
    ans += get(1, 1, n, i + 1, n, a[i].first);
  }
  cout << ans;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 N = 200005;
int n;
long long ans;
pair<long long, long long> a[N];
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  if (x.second != y.second) {
    return x.second < y.second;
  } else {
    return x.first < y.first;
  }
}
vector<long long> segmenttree[4 * N], segmenttree2[4 * N];
void build(int id, int l, int r) {
  if (l == r) {
    segmenttree[id] = {a[l].first};
    segmenttree2[id] = {a[l].first};
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  merge(segmenttree[id << 1].begin(), segmenttree[id << 1].end(),
        segmenttree[id << 1 | 1].begin(), segmenttree[id << 1 | 1].end(),
        back_inserter(segmenttree[id]));
  segmenttree2[id].resize(r - l + 1);
  segmenttree2[id][r - l] = segmenttree[id][r - l];
  for (int i = r - l - 1; i >= 0; i--) {
    segmenttree2[id][i] = segmenttree2[id][i + 1] + segmenttree[id][i];
  }
}
long long get(int id, int l, int r, int u, int v, long long val) {
  if (v < l || r < u) {
    return 0;
  }
  if (u <= l && r <= v) {
    long long idx =
        lower_bound(segmenttree[id].begin(), segmenttree[id].end(), val) -
        segmenttree[id].begin();
    if (idx == r - l + 1) {
      return 0;
    } else {
      return segmenttree2[id][idx] - (r - l + 1 - idx) * val;
    }
  }
  int mid = (l + r) >> 1;
  return get(id << 1, l, mid, u, v, val) +
         get(id << 1 | 1, mid + 1, r, u, v, val);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i].second;
  }
  sort(a + 1, a + 1 + n, cmp);
  build(1, 1, n);
  for (int i = 1; i < n; i++) {
    ans += get(1, 1, n, i + 1, n, a[i].first);
  }
  cout << ans;
}

```