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 maxn = 2e5 + 7;
struct node {
  long long val;
  long long time;
} a[maxn];
bool cmp(node aa, node bb) {
  if (aa.time == bb.time) {
    return aa.val < bb.val;
  } else {
    return aa.time > bb.time;
  }
}
map<long long, long long> mp;
long long Find(int x) {
  if (mp[x] == 0)
    return x;
  else
    return mp[x] = Find(mp[x]);
}
void Merge(long long x, long long y) {
  long long t1 = Find(x);
  long long t2 = Find(y);
  if (t1 != t2) {
    mp[t1] = t2;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  mp.clear();
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].val);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i].time);
  }
  sort(a + 1, a + n + 1, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long res = Find(a[i].val);
    Merge(res, res + 1);
    ans += a[i].time * (res - a[i].val);
  }
  printf("%lld\n", ans);
  return 0;
}
