#include <bits/stdc++.h>
const int N = 2e5 + 1e4;
int n, cnt;
int mp[N], a[N], t[N];
struct node {
  int id, val;
} tmp_a[N];
struct node2 {
  int val;
  bool operator<(const node2 _b) const { return this->val < _b.val; }
};
bool cmp(node _a, node _b) { return _a.val < _b.val; }
std::vector<int> sta[N];
std::priority_queue<node2> q;
void init() {
  std::sort(tmp_a + 1, tmp_a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (i == 1 || tmp_a[i].val != tmp_a[i - 1].val) {
      cnt++;
      mp[cnt] = tmp_a[i].val;
    }
    a[tmp_a[i].id] = cnt;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tmp_a[i].val);
    tmp_a[i].id = i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t[i]);
  }
  init();
  for (int i = 1; i <= n; i++) {
    sta[a[i]].push_back(i);
  }
  int la = 0;
  long long cur_sta = mp[1], ans = 0, sum = 0;
  while (1) {
    if (cur_sta == mp[la + 1]) {
      la++;
      int size = sta[la].size();
      for (int i = 0; i < size; i++) {
        int val = t[sta[la][i]];
        q.push((node2){val});
        sum += val;
      }
    }
    if (q.empty()) {
      if (la == cnt) break;
      cur_sta = mp[la + 1];
      continue;
    }
    int top = q.top().val;
    sum -= top;
    q.pop();
    ans += sum;
    cur_sta++;
  }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write a cpp solution to 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>
const int N = 2e5 + 1e4;
int n, cnt;
int mp[N], a[N], t[N];
struct node {
  int id, val;
} tmp_a[N];
struct node2 {
  int val;
  bool operator<(const node2 _b) const { return this->val < _b.val; }
};
bool cmp(node _a, node _b) { return _a.val < _b.val; }
std::vector<int> sta[N];
std::priority_queue<node2> q;
void init() {
  std::sort(tmp_a + 1, tmp_a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (i == 1 || tmp_a[i].val != tmp_a[i - 1].val) {
      cnt++;
      mp[cnt] = tmp_a[i].val;
    }
    a[tmp_a[i].id] = cnt;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tmp_a[i].val);
    tmp_a[i].id = i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t[i]);
  }
  init();
  for (int i = 1; i <= n; i++) {
    sta[a[i]].push_back(i);
  }
  int la = 0;
  long long cur_sta = mp[1], ans = 0, sum = 0;
  while (1) {
    if (cur_sta == mp[la + 1]) {
      la++;
      int size = sta[la].size();
      for (int i = 0; i < size; i++) {
        int val = t[sta[la][i]];
        q.push((node2){val});
        sum += val;
      }
    }
    if (q.empty()) {
      if (la == cnt) break;
      cur_sta = mp[la + 1];
      continue;
    }
    int top = q.top().val;
    sum -= top;
    q.pop();
    ans += sum;
    cur_sta++;
  }
  printf("%lld\n", ans);
}

```