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 double eps = 1e-6;
const int mod = 998244353;
const int maxn = 2e6 + 100;
const int maxm = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
int t;
int n;
struct node {
  int x, y;
} a[maxn];
map<int, int> ma;
multiset<int> v[maxn];
multiset<int> s;
bool cmp(node a, node b) { return a.x < b.x; }
long long S[maxn];
int X[maxn];
int main() {
  long long ans = 0;
  scanf("%d", &n);
  int tot = 0;
  int bg = 0x3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].x);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i].y);
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = 1; i <= n; i++) {
    bg = min(bg, a[i].x);
    if (!ma[a[i].x]) {
      ma[a[i].x] = ++tot;
      X[tot] = a[i].x;
    }
    v[ma[a[i].x]].insert(a[i].y);
    S[ma[a[i].x]] += a[i].y;
  }
  int now = bg;
  long long sum = 0;
  for (int i = 1; i <= tot; i++) {
    int x = X[i];
    while (now < x) {
      if (s.empty()) {
        now = x;
        break;
      }
      auto it = s.end();
      it--;
      s.erase(it);
      sum -= (*it);
      ans += sum;
      now++;
    }
    if (x == now) {
      if (v[i].size() > s.size()) {
        v[i].swap(s);
        sum = S[i];
      }
      for (auto it = v[i].begin(); it != v[i].end(); it++) {
        s.insert(*it);
        sum += (*it);
      }
      v[i].clear();
      auto it = s.end();
      it--;
      sum -= (*it);
      s.erase(it);
      ans += sum;
      now++;
      continue;
    }
  }
  while (!s.empty()) {
    auto it = s.end();
    it--;
    sum -= (*it);
    s.erase(it);
    ans += sum;
    now++;
  }
  printf("%lld", ans);
  return 0;
}
