#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const long double eps = 1e-8;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const int MOD = 1e9 + 7;
mt19937 rnd(239);
const int M = (int)(2e5 + 239);
const int N = (int)(2e3 + 239);
const int L = 20;
const int T = (1 << 18) + 239;
const int B = 500;
const int X = 5010;
int n, a[M], t[M];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  vector<pair<int, int>> al;
  for (int i = 0; i < n; i++) al.push_back(make_pair(a[i], i));
  sort(al.begin(), al.end());
  set<pair<int, int>> q;
  long long ans = 0;
  int p = 0;
  int ct = -1;
  long long st = 0;
  while (!q.empty() || p < n) {
    if (q.empty()) ct = al[p].first;
    while (p < n && ct == al[p].first) {
      int i = al[p].second;
      p++;
      st += t[i];
      q.insert(make_pair(t[i], i));
    }
    ans += st;
    auto it = q.end();
    it--;
    ans -= it->first;
    st -= it->first;
    q.erase(it);
    ct++;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int alf = 26;
const int dig = 10;
const int two = 2;
const int th = 3;
const long long prost = 239;
const long long btc = 30;
const long double eps = 1e-8;
const long long INF = (long long)(1e18 + 239);
const int BIG = (int)(1e9 + 239);
const int MOD = 1e9 + 7;
mt19937 rnd(239);
const int M = (int)(2e5 + 239);
const int N = (int)(2e3 + 239);
const int L = 20;
const int T = (1 << 18) + 239;
const int B = 500;
const int X = 5010;
int n, a[M], t[M];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> t[i];
  vector<pair<int, int>> al;
  for (int i = 0; i < n; i++) al.push_back(make_pair(a[i], i));
  sort(al.begin(), al.end());
  set<pair<int, int>> q;
  long long ans = 0;
  int p = 0;
  int ct = -1;
  long long st = 0;
  while (!q.empty() || p < n) {
    if (q.empty()) ct = al[p].first;
    while (p < n && ct == al[p].first) {
      int i = al[p].second;
      p++;
      st += t[i];
      q.insert(make_pair(t[i], i));
    }
    ans += st;
    auto it = q.end();
    it--;
    ans -= it->first;
    st -= it->first;
    q.erase(it);
    ct++;
  }
  cout << ans << "\n";
  return 0;
}

```