#include <bits/stdc++.h>
using namespace std;
const int inf = 2e9 + 7;
const long long Inf = 1e18 + 7;
const int maxn = 2e5 + 5;
const long long mod = 998244353;
int n;
int a[maxn], t[maxn];
struct cmp {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    return a.second < b.second;
  }
};
priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> q;
map<int, int> fa;
int find(int x) { return fa[x] == 0 ? x : fa[x] = find(fa[x]); }
void ba(int x, int y) {
  int px = find(x), py = find(y);
  if (px > py) {
    fa[py] = px;
  } else {
    fa[px] = py;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    q.push(pair<int, int>(a[i], t[i]));
  }
  long long ans = 0;
  while (!q.empty()) {
    pair<int, int> p = q.top();
    q.pop();
    ans += 1ll * (find(p.first) - p.first) * p.second;
    ba(find(p.first), find(p.first) + 1);
  }
  cout << ans << endl;
}

### Prompt

Generate 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>
using namespace std;
const int inf = 2e9 + 7;
const long long Inf = 1e18 + 7;
const int maxn = 2e5 + 5;
const long long mod = 998244353;
int n;
int a[maxn], t[maxn];
struct cmp {
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    return a.second < b.second;
  }
};
priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> q;
map<int, int> fa;
int find(int x) { return fa[x] == 0 ? x : fa[x] = find(fa[x]); }
void ba(int x, int y) {
  int px = find(x), py = find(y);
  if (px > py) {
    fa[py] = px;
  } else {
    fa[px] = py;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    q.push(pair<int, int>(a[i], t[i]));
  }
  long long ans = 0;
  while (!q.empty()) {
    pair<int, int> p = q.top();
    q.pop();
    ans += 1ll * (find(p.first) - p.first) * p.second;
    ba(find(p.first), find(p.first) + 1);
  }
  cout << ans << endl;
}

```