VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects a_i publications.

The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within t_i seconds. 

What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can't remove publications recommended by the batch algorithm.

Input

The first line of input consists of single integer n — the number of news categories (1 ≤ n ≤ 200 000).

The second line of input consists of n integers a_i — the number of publications of i-th category selected by the batch algorithm (1 ≤ a_i ≤ 10^9).

The third line of input consists of n integers t_i — time it takes for targeted algorithm to find one new publication of category i (1 ≤ t_i ≤ 10^5).

Output

Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.

Examples

Input


5
3 7 9 7 8
5 2 5 7 5


Output


6


Input


5
1 2 3 4 5
1 1 1 1 1


Output


0

Note

In the first example, it is possible to find three publications of the second type, which will take 6 seconds.

In the second example, all news categories contain a different number of publications.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  char ch = getchar();
  bool flag = false;
  x = 0;
  while (!isdigit(ch)) flag |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  if (flag) x = -x;
}
const int inf = 0x3f3f3f3f;
const double eps = 10e-5;
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
map<int, int> fa;
int Find(int x) { return fa[x] == 0 ? x : fa[x] = Find(fa[x]); }
void Union(int x, int y) {
  x = Find(x), y = Find(y);
  fa[x] = y;
}
bool cmp(pair<int, int> &a, pair<int, int> &b) {
  return a.second > b.second || (a.second == b.second && a.first > b.first);
}
int n;
pair<int, int> a[N];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i].first);
  for (int i = 1; i <= n; i++) read(a[i].second);
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long fx = Find(a[i].first);
    ans += (fx - a[i].first) * a[i].second;
    Union(fx, fx + 1);
  }
  cout << ans << endl;
  return 0;
}
