#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-5;
int n;
pair<long long, long long> a[maxn];
long long b[maxn];
long long s1[maxn], s2[maxn];
int lb(int x) { return x & (-x); }
void add(int x, long long w) {
  for (int i = x; i <= n; i += lb(i)) {
    s1[i]++;
    s2[i] += w;
  }
}
long long sum1(int x) {
  long long res = 0;
  for (int i = x; i >= 1; i -= lb(i)) res += s1[i];
  return res;
}
long long sum2(int x) {
  long long res = 0;
  for (int i = x; i >= 1; i -= lb(i)) res += s2[i];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].first);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].second);
    b[i] = a[i].second;
  }
  sort(a + 1, a + 1 + n);
  sort(b + 1, b + 1 + n);
  int len = unique(b + 1, b + 1 + n) - (b + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(b + 1, b + 1 + len, a[i].second) - b;
    long long res1 = sum1(pos);
    long long res2 = sum2(pos);
    ans += res1 * a[i].first - res2;
    add(pos, a[i].first);
  }
  printf("%lld", ans);
}

### Prompt

Develop a solution in cpp 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 maxn = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-5;
int n;
pair<long long, long long> a[maxn];
long long b[maxn];
long long s1[maxn], s2[maxn];
int lb(int x) { return x & (-x); }
void add(int x, long long w) {
  for (int i = x; i <= n; i += lb(i)) {
    s1[i]++;
    s2[i] += w;
  }
}
long long sum1(int x) {
  long long res = 0;
  for (int i = x; i >= 1; i -= lb(i)) res += s1[i];
  return res;
}
long long sum2(int x) {
  long long res = 0;
  for (int i = x; i >= 1; i -= lb(i)) res += s2[i];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].first);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].second);
    b[i] = a[i].second;
  }
  sort(a + 1, a + 1 + n);
  sort(b + 1, b + 1 + n);
  int len = unique(b + 1, b + 1 + n) - (b + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int pos = lower_bound(b + 1, b + 1 + len, a[i].second) - b;
    long long res1 = sum1(pos);
    long long res2 = sum2(pos);
    ans += res1 * a[i].first - res2;
    add(pos, a[i].first);
  }
  printf("%lld", ans);
}

```