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>
const double eps = (1e-9);
using namespace std;
int dcmp(long double a, long double b) {
  return fabsl(a - b) <= eps ? 0 : (a > b) ? 1 : -1;
}
int getBit(long long num, int idx) { return ((num >> idx) & 1ll) == 1; }
int setBit1(long long num, long long idx) { return num | (1ll << idx); }
long long setBit0(long long num, int idx) { return num & ~(1ll << idx); }
long long flipBit(long long num, int idx) { return num ^ (1ll << idx); }
void run() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int countNumBit1(int mask) {
  int ret = 0;
  while (mask) {
    mask &= (mask - 1);
    ++ret;
  }
  return ret;
}
const int N = 200000 + 9;
map<int, priority_queue<long long> > mb;
map<int, long long> sum;
int arr[N], t[N];
set<int> st2;
set<pair<long long, long long> > st;
int main() {
  run();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    mb[arr[i]].push(t[i]);
    sum[arr[i]] += t[i];
  }
  int str = INT_MAX;
  for (auto it = mb.begin(); it != mb.end(); it++) {
    if (((int)((*it).second).size()) > 1) {
      st2.insert((*it).first);
    }
  }
  long long res = 0;
  while (((int)(st2).size()) > 0) {
    int str = (*st2.begin());
    st.clear();
    st.insert(make_pair(-mb[str].top(), str));
    long long gsum = sum[str];
    int cnt = ((int)(mb[str]).size());
    while (1) {
      auto it = st.begin();
      long long ff = -(*it).first;
      long long ss = (*it).second;
      mb[ss].pop();
      sum[ss] -= ff;
      gsum -= ff;
      res += gsum;
      st.erase(st.begin());
      if (((int)(mb[ss]).size()) > 0) st.insert(make_pair(-mb[ss].top(), ss));
      if (((int)(mb[str + 1]).size()) > 0)
        st.insert(make_pair(-mb[str + 1].top(), str + 1));
      st2.erase(str);
      cnt--;
      cnt += ((int)(mb[str + 1]).size());
      if (cnt == 1) break;
      gsum += sum[str + 1];
      str++;
    }
  }
  cout << res << endl;
}
