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.
//package com.example.demo_Amdocs;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;

public class GFG {

  static class Pair {

    int x;
    int y;

    Pair(int x, int y) {
      this.x = x;
      this.y = y;
    }
  }

  private static int max(ArrayList<Integer> A) {
    int max = -1, ind = -1;
    ;
    for (int i = 0; i < A.size(); i++) {
      if (A.get(i) > max) {
        max = A.get(i);
        ind = i;
      }
    }
    return ind;
  }

  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //int t=Integer.parseInt(br.readLine().trim());
    //while(t--!=0){
    int n = Integer.parseInt(br.readLine().trim());
    String s = br.readLine().trim();
    String ss[] = s.split("\\s+");
    int a[] = new int[n];
    for (int i = 0; i < n; i++) {
      a[i] = Integer.parseInt(ss[i]);
    }
    s = br.readLine().trim();
    ss = s.split("\\s+");
    int t[] = new int[n];
    for (int i = 0; i < n; i++) {
      t[i] = Integer.parseInt(ss[i]);
    }
    Pair[] p = new Pair[n];
    for (int i = 0; i < n; i++) {
      p[i] = new Pair(a[i], t[i]);
    }
    Arrays.sort(p, (p1, p2) -> p1.x - p2.x);
    long cnt=0;
    PriorityQueue<Pair> q = new PriorityQueue<>((p1, p2) -> p2.y - p1.y);
    int val = p[0].x;
    for (int i = 0; i < n; i++) {
      if (val == p[i].x) {
        q.add(p[i]);
        continue;
      }
      while (!q.isEmpty()&&val<p[i].x) {
        Pair o=q.poll();
        cnt+=1l*(val-o.x)*o.y;
        val++;
      }
      val=p[i].x;
      q.add(p[i]);
    }
    while (!q.isEmpty()) {
      Pair o=q.poll();
      cnt+=1l*(val-o.x)*o.y;
      val++;
    }
    System.out.println(cnt);
    //}
  }
}
	