#include <bits/stdc++.h>
using namespace std;
struct node {
  int x, t;
} a[200010];
unordered_map<int, int> fa;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline bool cmp(const node &x, const node &y) {
  if (x.t > y.t)
    return true;
  else if (x.t == y.t && x.x < y.x)
    return true;
  return false;
}
inline int findf(int x) { return fa[x] == 0 ? x : fa[x] = findf(fa[x]); }
void merge(int x, int y) {
  int X = findf(x), Y = findf(y);
  if (X != Y) fa[X] = Y;
  return;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) a[i].x = read();
  for (int i = 1; i <= n; i++) a[i].t = read();
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int f = findf(a[i].x);
    if (f != a[i].x) ans += 1ll * (f - a[i].x) * a[i].t;
    merge(f, f + 1);
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Please formulate 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;
struct node {
  int x, t;
} a[200010];
unordered_map<int, int> fa;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline bool cmp(const node &x, const node &y) {
  if (x.t > y.t)
    return true;
  else if (x.t == y.t && x.x < y.x)
    return true;
  return false;
}
inline int findf(int x) { return fa[x] == 0 ? x : fa[x] = findf(fa[x]); }
void merge(int x, int y) {
  int X = findf(x), Y = findf(y);
  if (X != Y) fa[X] = Y;
  return;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) a[i].x = read();
  for (int i = 1; i <= n; i++) a[i].t = read();
  sort(a + 1, a + 1 + n, cmp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int f = findf(a[i].x);
    if (f != a[i].x) ans += 1ll * (f - a[i].x) * a[i].t;
    merge(f, f + 1);
  }
  printf("%lld", ans);
  return 0;
}

```