CF1027D Mouse Hunt 思维

Mouse Hunt

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

The dormitory consists of $\mathrm{nn rooms\; and\; a\; single\; mouse!\; Girls\; decided\; to\; set\; mouse\; traps\; in\; some\; rooms\; to\; get\; rid\; of\; the\; horrible\; monster.\; Setting\; a\; trap\; in\; room\; number ii costs cici burles.\; Rooms\; are\; numbered\; from 11 to nn.}$

Mouse doesn't sit in place all the time, it constantly runs. If it is in room $\mathrm{ii in\; second tt then\; it\; will\; run\; to\; room aiai in\; second t+1t+1 without\; visiting\; any\; other\; rooms\; inbetween\; (i=aii=ai means\; that\; mouse\; won\text{'}t\; leave\; room ii).\; It\text{'}s\; second 00 in\; the\; start.\; If\; the\; mouse\; is\; in\; some\; room\; with\; a\; mouse\; trap\; in\; it,\; then\; the\; mouse\; get\; caught\; into\; this\; trap.}$

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 $\mathrm{11 to nn at\; second 00.}$

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 $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; number\; of\; rooms\; in\; the\; dormitory.}$

The second line contains $\mathrm{nn integers c1,c2,\dots ,cnc1,c2,\dots ,cn (1\le ci\le 1041\le ci\le 104)\; — cici is\; the\; cost\; of\; setting\; the\; trap\; in\; room\; number ii.}$

The third line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le n1\le ai\le n)\; — aiai is\; the\; room\; the\; mouse\; will\; run\; to\; the\; next\; second\; after\; being\; in\; room ii.}$

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

input

Copy

5
1 2 3 2 10
1 3 4 3 3

output

Copy

3

input

Copy

4
1 10 2 10
2 4 2 2

output

Copy

10

input

Copy

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

output

Copy

2

Note

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

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

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

• $\mathrm{1\to 2\to 2\to \dots 1\to 2\to 2\to \dots ;}$
• $\mathrm{2\to 2\to \dots 2\to 2\to \dots ;}$
• $\mathrm{3\to 2\to 2\to \dots 3\to 2\to 2\to \dots ;}$
• $\mathrm{4\to 3\to 2\to 2\to \dots 4\to 3\to 2\to 2\to \dots ;}$
• $\mathrm{5\to 6\to 7\to 6\to \dots 5\to 6\to 7\to 6\to \dots ;}$
• $\mathrm{6\to 7\to 6\to \dots 6\to 7\to 6\to \dots ;}$
• $\mathrm{7\to 6\to 7\to \dots 7\to 6\to 7\to \dots ;}$

So it's enough to set traps in rooms $\mathrm{22 and 66.}$

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 1e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll n, ans, a[maxn], c[maxn], vis[maxn];
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n;
    for( ll i = 1; i <= n; i ++ ) {
        cin >> c[i];
    }
    for( ll i = 1; i <= n; i ++ ) {
        cin >> a[i];
    }
    for( ll i = 1; i <= n; i ++ ) {
        ll x = i, v, mn;
        while( !vis[x] ) {  // 找到已被访问过的下一个房间
            vis[x] = i, x = a[x];
        }
        if( vis[x] != i ) {
            continue; 
        }
        v = x, mn = c[x];
        while( a[x] != v ) { //在可能经过的所有房间需要花费最小的房间下陷阱
            x = a[x], mn = min( mn, c[x] );
        }
        ans += mn;
    }
    cout << ans << endl;
    return 0;
}