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("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long mod1 = 998244353;
const long long mod = 1e9 + 7;
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pwr(long long a, long long b) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long findMinNumber(long long n) {
  long long count = 0, ans = 1;
  while (n % 2 == 0) {
    count++;
    n /= 2;
  }
  if (count % 2) ans *= 2;
  for (int i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n /= i;
    }
    if (count % 2) ans *= i;
  }
  if (n > 2) ans *= n;
  return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n], b[n], i, sum = 0, j;
  vector<pair<long long, long long> > v(n);
  set<long long> s;
  map<long long, long long> m;
  long long cnt = 0, nextpos = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
    if (m[a[i]] == 1) cnt++;
    s.insert(a[i]);
  }
  for (i = 0; i < n; i++) {
    cin >> b[i];
    v[i] = {b[i], a[i]};
  }
  sort(a, a + n);
  for (i = 0; i < n; i++) {
    for (j = max(a[i] + 1, nextpos); j < 1e10; j++) {
      if (m[j] == 0) {
        s.insert(j);
        m[j]++;
        nextpos = j + 1;
        break;
      }
    }
  }
  if (cnt == n)
    cout << 0 << '\n';
  else {
    sort((v).begin(), (v).end());
    for (i = n - 1; i >= 0; i--) {
      auto it = s.lower_bound(v[i].second);
      sum += (*it - v[i].second) * v[i].first;
      s.erase(it);
    }
    cout << sum << '\n';
  }
  return 0;
}
