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;
const int Inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 2e5 + 50;
struct N {
  int v, p;
  N() {}
  N(int v1, int p1) : v(v1), p(p1) {}
} tmp[maxn];
struct cmp1 {
  bool operator()(const N& a, const N& b) {
    if (a.v != b.v) return a.v > b.v;
    return a.p < b.p;
  }
};
bool cmp(const N& a, const N& b) { return a.v < b.v; }
priority_queue<int> que;
N a[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].v);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].p);
  }
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0, sum = 0, cur = 0;
  for (int i = 1; i <= n; i++) {
    cur = a[i].v;
    sum = a[i].p;
    que.push(a[i].p);
    int j = i + 1;
    while (!que.empty()) {
      while (j <= n && a[j].v <= cur) {
        que.push(a[j].p);
        sum += a[j].p;
        j++;
      }
      sum -= que.top();
      que.pop();
      ans += sum;
      cur++;
    }
    i = j - 1;
  }
  cout << ans << endl;
}
