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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
using LL = int_fast64_t;
const int MAXN = 2e5 + 7;
int n, A[MAXN], B[MAXN];
map<int, int> msk;
struct LTT {
  int dist[MAXN << 2], ls[MAXN << 2], rs[MAXN << 2], val[MAXN << 2], ID;
  unordered_map<int, LL> sum;
  LTT() {
    dist[0] = -1;
    ID = 0;
  }
  int merge(int x, int y) {
    if (!x || !y) return x | y;
    if (val[x] < val[y]) swap(x, y);
    rs[x] = merge(rs[x], y);
    if (dist[ls[x]] < dist[rs[x]]) swap(ls[x], rs[x]);
    dist[x] = dist[rs[x]] + 1;
    return x;
  }
  void push(int x, int y) {
    if (!msk.count(x)) {
      msk.insert(make_pair(x, ++ID));
      val[ID] = y;
      sum[x] = y;
    } else {
      val[++ID] = y;
      msk[x] = merge(msk.at(x), ID);
      sum[x] += y;
    }
  }
  void pop(int x) {
    sum[x] -= val[msk.at(x)];
    msk.at(x) = merge(ls[msk.at(x)], rs[msk.at(x)]);
  }
  bool empty(int x) { return !msk.at(x); }
} ltt;
int main() {
  ____();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= n; i++) cin >> B[i];
  for (int i = 1; i <= n; i++) ltt.push(A[i], B[i]);
  int cur = msk.begin()->first;
  LL cost = 0;
  while (true) {
    if (!msk.count(cur)) {
      auto p = msk.upper_bound(cur);
      if (p == msk.end()) break;
      cur = p->first;
    }
    ltt.pop(cur);
    cost += ltt.sum[cur];
    if (ltt.empty(cur)) {
      msk.erase(cur);
      continue;
    }
    ltt.sum[cur + 1] += ltt.sum[cur];
    if (!msk.count(cur + 1)) {
      msk.insert(make_pair(cur + 1, msk.at(cur)));
      msk.erase(cur);
    } else
      msk.at(cur + 1) = ltt.merge(msk.at(cur + 1), msk.at(cur));
    cur++;
  }
  cout << cost << endl;
  return 0;
}
