Medicine faculty of Berland State University has just finished their admission campaign. As usual, about 80%80% of applicants are girls and majority of them are going to live in the university dormitory for the next 44 (hopefully) years.

The dormitory consists of nn 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 ii costs cici burles. Rooms are numbered from 11 to nn.

Mouse doesn't sit in place all the time, it constantly runs. If it is in room ii in second tt then it will run to room aiai in second t+1t+1 without visiting any other rooms inbetween (i=aii=ai means that mouse won't leave room ii). It's second 00 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 11 to nn at second 00.

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 nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of rooms in the dormitory.

The second line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1041≤ci≤104) — cici is the cost of setting the trap in room number ii.

The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — aiai is the room the mouse will run to the next second after being in room ii.

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

5
1 2 3 2 10
1 3 4 3 3

3

4
1 10 2 10
2 4 2 2

10

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

2

Note

In the first example it is enough to set mouse trap in rooms 11 and 44. If mouse starts in room 11 then it gets caught immideately. If mouse starts in any other room then it eventually comes to room 44.

In the second example it is enough to set mouse trap in room 22. If mouse starts in room 22 then it gets caught immideately. If mouse starts in any other room then it runs to room 22 in second 11.

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

• 1→2→2→…1→2→2→…;
• 2→2→…2→2→…;
• 3→2→2→…3→2→2→…;
• 4→3→2→2→…4→3→2→2→…;
• 5→6→7→6→…5→6→7→6→…;
• 6→7→6→…6→7→6→…;
• 7→6→7→…7→6→7→…;

So it's enough to set traps in rooms 22 and 66.

一个连通块 肯定存在一个环

所以找环上的最小值即可

为什么一个连通块肯定存在一个环  因为每个点都有一个出度 。。。

#include <bits/stdc++.h>

#define mem(a, b) memset(a, b, sizeof(a))

using namespace std;

const int maxn = 1e5 + , INF = 0x7fffffff;

typedef long long LL;

int n, cnt;

int a[maxn<<], head[maxn<<], vis[maxn<<], pre[maxn<<];

int s, t;

struct node

{

    int u, v, next;

}Node[maxn<<];

void add(int u, int v)

{

    Node[cnt].u = u;

    Node[cnt].v = v;

    Node[cnt].next = head[u];

    head[u] = cnt++;

}

void dfs1(int u, int fa)

{

    vis[u] = ;

    for(int i=head[u]; i!=-; i=Node[i].next)

    {

        node e = Node[i];

        if(!vis[e.v])

        {

            pre[e.v] = u;

            dfs1(e.v, u);

        }

        else

        {

            s = e.v;

            t = u;

            return;

        }

    }

}

int minn = INF;

LL res = ;

void dfs2(int u)    //由t向e.v回溯，如果能回溯到s则说明这是一个新的环  那么就把res += minn   其实放到一个dfs里就好了

{

    minn = min(minn, a[u]);

    if(u == s)

    {

        res += (LL)minn;

        return;

    }

    else if(u == ) return;

    dfs2(pre[u]);

}

int main()

{

    mem(head, -);

    cnt = ;

    cin>> n;

    for(int i=; i<=n; i++)

        cin>> a[i];

    int u;

    for(int i=; i<=n; i++)

    {

        cin>> u;

        add(i, u);

    }

    for(int i=; i<=n; i++)

    {

        if(!vis[i])

        {

            minn = INF;

            d[i] = ;

            dfs1(i, -);

            dfs2(t);

        }

    }

    cout<< res <<endl;

    return ;

}