Medicine faculty of Berland State University has just finished their admission campaign. As usual, about 80\% of applicants are girls and majority of them are going to live in the university dormitory for the next 4 (hopefully) years.

The dormitory consists of n rooms and a single mouse! Girls decided to set mouse traps in some rooms to get rid of the horrible monster. Setting a trap in room number i costs c_i burles. Rooms are numbered from 1 to n.

Mouse doesn't sit in place all the time, it constantly runs. If it is in room i in second t then it will run to room a_i in second t + 1 without visiting any other rooms inbetween (i = a_i means that mouse won't leave room i). It's second 0 in the start. If the mouse is in some room with a mouse trap in it, then the mouse get caught into this trap.

That would have been so easy if the girls actually knew where the mouse at. Unfortunately, that's not the case, mouse can be in any room from 1 to n at second 0.

What it the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from?

Input

The first line contains as single integers n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of rooms in the dormitory.

The second line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^4) — c_i is the cost of setting the trap in room number i.

The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — a_i is the room the mouse will run to the next second after being in room i.

Output

Print a single integer — the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from.

Examples

Input

5
1 2 3 2 10
1 3 4 3 3


Output

3


Input

4
1 10 2 10
2 4 2 2


Output

10


Input

7
1 1 1 1 1 1 1
2 2 2 3 6 7 6


Output

2

Note

In the first example it is enough to set mouse trap in rooms 1 and 4. If mouse starts in room 1 then it gets caught immideately. If mouse starts in any other room then it eventually comes to room 4.

In the second example it is enough to set mouse trap in room 2. If mouse starts in room 2 then it gets caught immideately. If mouse starts in any other room then it runs to room 2 in second 1.

Here are the paths of the mouse from different starts from the third example:

  * 1 → 2 → 2 → ...; 
  * 2 → 2 → ...; 
  * 3 → 2 → 2 → ...; 
  * 4 → 3 → 2 → 2 → ...; 
  * 5 → 6 → 7 → 6 → ...; 
  * 6 → 7 → 6 → ...; 
  * 7 → 6 → 7 → ...; 



So it's enough to set traps in rooms 2 and 6.
3
1 2 4
2 2 1
 5
5 4 2 6 10
2 4 3 3 2
 4
2 1 2 2
2 3 3 2
 8
1 2 2 9 7 5 6 5
8 4 1 5 5 1 1 8
 3
76 74 84
3 2 1
 3
1 2 4
2 3 3
 7
2 1 64 8 32 4 16
5 7 2 4 2 4 6
 3
1 2 4
3 2 3
 5
1 1 10 10 10
2 3 4 5 3
 4
1 12 2 3
2 3 4 2
 4
99 71 47 82
1 3 4 2
 5
1 2 3 4 5
3 4 5 3 2
 4
90 5 66 84
3 1 2 4
 3
9 5 6
3 3 1
 8
8 8 1 10 6 9 3 1
5 2 7 1 6 8 4 3
 3
80 6 67
1 2 3
 10
10 10 10 10 1 10 10 10 10 10
2 3 4 5 3 2 6 9 5 8
 10
6870 3547 3337 9426 3866 1873 4908 4628 7800 6716
2 3 4 5 1 4 6 2 8 9
 4
97 27 70 99
1 2 3 4
 4
2 58 52 89
3 1 4 2
 3
1 2 4
1 3 1
 3
1 2 4
3 3 3
 4
81 69 80 19
4 3 1 2
 4
4 95 20 62
2 1 4 3
 2
62 49
1 2
 4
1 2 4 8
2 3 3 2
 8
1 8 32 4 16 64 128 2
6 4 1 5 4 4 1 3
 2
21 28
2 1
 10
5620 2478 4051 7060 6327 2170 6912 2031 1781 7454
1 1 1 1 1 1 1 1 1 1
 6
10 7 1 1 8 2
5 5 2 3 3 5
 2
1 2
2 2
 10
6 9 1 1 1 10 2 4 9 6
5 3 4 2 6 8 9 1 10 7
 3
15 52 14
1 3 2
 1
12
1
 5
1 16 2 4 8
5 3 1 5 4
 4
88 78 39 46
4 3 2 1
 3
1 2 4
3 3 1
 3
1 2 4
3 2 2
 4
37 86 1 88
4 2 1 3
 10
6724 5003 9029 376 7529 5554 1407 388 2632 981
2 3 4 5 1 4 6 7 8 9
 10
1337 4359 5138 5246 7859 8111 1154 8455 3711 1737
1 1 2 3 4 1 6 7 8 9
 3
40 26 88
3 1 2
 10
7846 8354 1188 6898 8046 7561 8043 8588 8684 3699
1 1 2 3 4 5 6 7 8 9
 7
4 32 8 1 16 64 2
6 1 5 2 1 4 5
 3
30 98 37
2 3 1
 3
17 34 88
2 1 3
 10
9 19 19 1 3 9 1 12 10 8
7 1 6 3 4 8 5 10 2 9

2 2 2 12 150 4 8 6 10 2 146 2 89 6 9 153 1 3337 293 2 1 4 19 24 111 4 4 21 5620 1 2 4 29 12 4 85 1 2 87 376 1337 26 7846 1 30 105 1
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int c[maxn], a[maxn], vis[maxn];
int sav = 0, dp[maxn];
bool c1 = false;
void dfs(int u) {
  if (vis[u]) return;
  vis[u] = true;
  if (vis[a[u]] && !c1) {
    sav = c[u], c1 = true;
    return;
  }
  dfs(a[u]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n; i++) cin >> a[i];
  long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      if (vis[a[i]]) {
        vis[i] = true;
        continue;
      }
      c1 = false;
      sav = 0;
      dfs(i);
      sum += sav;
    }
  }
  cout << sum;
  return 0;
}
