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.
#include <bits/stdc++.h>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (c == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
struct SkewHeapNode {
  int value;
  SkewHeapNode *left;
  SkewHeapNode *right;
  SkewHeapNode() {}
  SkewHeapNode(int a) {
    this->value = a;
    this->left = NULL;
    this->right = NULL;
  }
};
struct SkewHeap {
  SkewHeapNode *root;
  int size, level;
  long long sum;
  SkewHeapNode *merge(SkewHeapNode *a, SkewHeapNode *b) {
    if (a == NULL) return b;
    if (b == NULL) return a;
    if (a->value > b->value) {
      a->right = merge(a->right, b);
      std::swap(a->left, a->right);
      return a;
    } else {
      b->right = merge(a, b->right);
      std::swap(b->left, b->right);
      return b;
    }
  }
  void merge(SkewHeap a) {
    this->size += a.size;
    this->sum += a.sum;
    this->root = merge(this->root, a.root);
  }
  void push(int a) {
    this->size++;
    this->sum += a;
    SkewHeapNode *tmp = new SkewHeapNode;
    *tmp = SkewHeapNode(a);
    this->root = merge(this->root, tmp);
  }
  void pop() {
    this->sum -= this->root->value;
    this->size--;
    SkewHeapNode *tmp = this->root;
    this->root = merge(this->root->left, this->root->right);
    delete tmp;
  }
} SkewHeap[200005];
int a[200005], u[200005];
inline void work() {
  long long n = read(), ans = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read(), u[i] = a[i];
  }
  std::sort(u + 1, u + n + 1);
  int cnt = std::unique(u + 1, u + n + 1) - u - 1;
  for (int i = 1; i <= n; i++) {
    int t = read();
    a[i] = std::lower_bound(u + 1, u + cnt + 1, a[i]) - u;
    SkewHeap[a[i]].push(t);
  }
  for (int i = 1; i <= cnt; i++) SkewHeap[i].level = u[i];
  SkewHeap[cnt + 1].level = 2e9;
  for (int i = 1; i <= cnt; i++) {
    for (; SkewHeap[i].size > 1 && SkewHeap[i].level < SkewHeap[i + 1].level;) {
      SkewHeap[i].pop();
      ans += SkewHeap[i].sum;
      SkewHeap[i].level++;
    }
    if (SkewHeap[i].level == SkewHeap[i + 1].level)
      SkewHeap[i + 1].merge(SkewHeap[i]);
  }
  printf("%lld\n", ans);
}
int main() {
  work();
  return 0;
}
