#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long linf = (1ll << 62) - 1;
const int N = 3e5 + 7;
const int mod = 1e9 + 7;
const double eps = 1e-12;
template <typename T>
inline void rd(T &x) {
  char c;
  int sign = 1;
  x = 0;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (c < '0' || c > '9');
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  x *= sign;
}
map<int, int> f;
int n, a[N], t[N];
long long b[N];
bool cmp(int i, int j) { return b[i] > b[j]; }
int get(int n) { return n == f[n] ? n : f[n] = get(f[n]); }
int main() {
  rd(n);
  for (int i = 1; i <= n; ++i) rd(a[i]);
  for (int i = 1; i <= n; ++i) rd(b[i]), t[i] = i;
  sort(t + 1, t + 1 + n, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int j = t[i];
    if (!f.count(a[j])) {
      f[a[j]] = a[j];
      if (f.count(a[j] - 1)) f[a[j] - 1] = a[j];
      if (f.count(a[j] + 1)) f[a[j]] = a[j] + 1;
    } else {
      int p = get(a[j]);
      ans += (long long)b[j] * (p - a[j] + 1);
      f[p] = p + 1, f[p + 1] = p + 1;
      if (f.count(p + 2)) f[p + 1] = p + 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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 = 0x3f3f3f3f;
const long long linf = (1ll << 62) - 1;
const int N = 3e5 + 7;
const int mod = 1e9 + 7;
const double eps = 1e-12;
template <typename T>
inline void rd(T &x) {
  char c;
  int sign = 1;
  x = 0;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (c < '0' || c > '9');
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  x *= sign;
}
map<int, int> f;
int n, a[N], t[N];
long long b[N];
bool cmp(int i, int j) { return b[i] > b[j]; }
int get(int n) { return n == f[n] ? n : f[n] = get(f[n]); }
int main() {
  rd(n);
  for (int i = 1; i <= n; ++i) rd(a[i]);
  for (int i = 1; i <= n; ++i) rd(b[i]), t[i] = i;
  sort(t + 1, t + 1 + n, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int j = t[i];
    if (!f.count(a[j])) {
      f[a[j]] = a[j];
      if (f.count(a[j] - 1)) f[a[j] - 1] = a[j];
      if (f.count(a[j] + 1)) f[a[j]] = a[j] + 1;
    } else {
      int p = get(a[j]);
      ans += (long long)b[j] * (p - a[j] + 1);
      f[p] = p + 1, f[p + 1] = p + 1;
      if (f.count(p + 2)) f[p + 1] = p + 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```