#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7, maxn = 2e5 + 10;
int n, m, a[maxn], ind;
long long ans, cnt;
priority_queue<int> q;
multiset<int>::iterator it;
struct P {
  int x, y;
  bool operator<(const P &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
} node[maxn];
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &node[i].x);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &node[i].y);
  }
  sort(node + 1, node + n + 1);
  node[0].x = node[0].y = 0;
  node[n + 1].x = node[n + 1].y = 0;
  int z = 0, r = 0;
  for (int i = 1; i <= n; ++i) {
    if (node[i].x == node[i + 1].x) {
      continue;
    }
    int t = q.size();
    t = min(t, node[i].x - z);
    z = node[i].x;
    while (t--) {
      cnt -= q.top();
      ans += cnt;
      q.pop();
    }
    for (int j = r + 1; j <= i; ++j) {
      q.push(node[j].y);
      cnt += node[j].y;
    }
    r = i;
  }
  while (!q.empty()) {
    cnt -= q.top();
    ans += cnt;
    q.pop();
  }
  cout << ans << endl;
  ;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int mod = 1e9 + 7, maxn = 2e5 + 10;
int n, m, a[maxn], ind;
long long ans, cnt;
priority_queue<int> q;
multiset<int>::iterator it;
struct P {
  int x, y;
  bool operator<(const P &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
} node[maxn];
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &node[i].x);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &node[i].y);
  }
  sort(node + 1, node + n + 1);
  node[0].x = node[0].y = 0;
  node[n + 1].x = node[n + 1].y = 0;
  int z = 0, r = 0;
  for (int i = 1; i <= n; ++i) {
    if (node[i].x == node[i + 1].x) {
      continue;
    }
    int t = q.size();
    t = min(t, node[i].x - z);
    z = node[i].x;
    while (t--) {
      cnt -= q.top();
      ans += cnt;
      q.pop();
    }
    for (int j = r + 1; j <= i; ++j) {
      q.push(node[j].y);
      cnt += node[j].y;
    }
    r = i;
  }
  while (!q.empty()) {
    cnt -= q.top();
    ans += cnt;
    q.pop();
  }
  cout << ans << endl;
  ;
  return 0;
}

```