#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
struct P {
  int a, t;
} p[MAXN];
bool cmp(P a, P b) { return a.t > b.t; }
int n, tot, rt, ch[MAXN << 5][2], val[MAXN << 5];
long long ans;
void modify(int &p, int l, int r, int pos, int v) {
  if (!p) {
    p = ++tot;
  }
  val[p] += v;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    modify(ch[p][0], l, mid, pos, v);
  } else {
    modify(ch[p][1], mid + 1, r, pos, v);
  }
  return;
}
int query(int p, int l, int r, int xl, int xr) {
  if (xr < l || r < xl) {
    return 0;
  }
  if (xl <= l && r <= xr) {
    return val[p];
  }
  int mid = (l + r) >> 1;
  return query(ch[p][0], l, mid, xl, xr) + query(ch[p][1], mid + 1, r, xl, xr);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].a);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].t);
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    int l = p[i].a, r = 1e9 + 1e6;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(rt, 1, 1e9 + 1e6, l, mid) == mid - l + 1) {
        l = mid + 1;
      } else {
        r = mid;
      }
    }
    ans += 1ll * p[i].t * (l - p[i].a);
    modify(rt, 1, 1e9 + 1e6, l, 1);
  }
  printf("%lld\n", ans);
  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 MAXN = 200010;
struct P {
  int a, t;
} p[MAXN];
bool cmp(P a, P b) { return a.t > b.t; }
int n, tot, rt, ch[MAXN << 5][2], val[MAXN << 5];
long long ans;
void modify(int &p, int l, int r, int pos, int v) {
  if (!p) {
    p = ++tot;
  }
  val[p] += v;
  if (l == r) {
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    modify(ch[p][0], l, mid, pos, v);
  } else {
    modify(ch[p][1], mid + 1, r, pos, v);
  }
  return;
}
int query(int p, int l, int r, int xl, int xr) {
  if (xr < l || r < xl) {
    return 0;
  }
  if (xl <= l && r <= xr) {
    return val[p];
  }
  int mid = (l + r) >> 1;
  return query(ch[p][0], l, mid, xl, xr) + query(ch[p][1], mid + 1, r, xl, xr);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].a);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].t);
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    int l = p[i].a, r = 1e9 + 1e6;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (query(rt, 1, 1e9 + 1e6, l, mid) == mid - l + 1) {
        l = mid + 1;
      } else {
        r = mid;
      }
    }
    ans += 1ll * p[i].t * (l - p[i].a);
    modify(rt, 1, 1e9 + 1e6, l, 1);
  }
  printf("%lld\n", ans);
  return 0;
}

```