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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 998244353;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 998244353;
    }
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 998244353 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a);
    }
    a = (a * a);
    b >>= 1LL;
  }
  return res;
}
const long long dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
const long long dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  set<long long> s;
  map<long long, vector<long long> > m;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    s.insert(v[i]);
  }
  for (long long i = 0; i < n; i++) {
    long long t;
    cin >> t;
    m[v[i]].push_back(t);
  }
  long long c = 0, ans = 0;
  priority_queue<long long> q;
  for (auto x : s) {
    for (auto y : m[x]) {
      c += y;
      q.push(y);
    }
    c -= (q.top());
    ans += c;
    q.pop();
    if (!q.empty()) s.insert(x + 1);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
