Directed Roads

ZS the Coder and Chris the Baboon has explored Udayland for quite some time. They realize that it consists of n towns numbered from1 to n.

There are n directed roads in the Udayland. i-th of them goes from town i to some other town ai (ai ≠ i). ZS the Coder can flip the direction of any road in Udayland, i.e. if it goes from town A to town B before the flip, it will go from town B to town A after.

ZS the Coder considers the roads in the Udayland confusing, if there is a sequence of distinct towns A1, A2, ..., Ak (k > 1) such that for every 1 ≤ i < k there is a road from town Ai to town Ai + 1 and another road from town Ak to town A1. In other words, the roads are confusing if some of them form a directed cycle of some towns.

Now ZS the Coder wonders how many sets of roads (there are 2n variants) in initial configuration can he choose to flip such that after flipping each road in the set exactly once, the resulting network will not be confusing.

Note that it is allowed that after the flipping there are more than one directed road from some town and possibly some towns with no roads leading out of it, or multiple roads between any pair of cities.

Input

The first line of the input contains single integer n (2 ≤ n ≤ 2·105) — the number of towns in Udayland.

The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n, ai ≠ i), ai denotes a road going from town i to town ai.

Output

Print a single integer — the number of ways to flip some set of the roads so that the resulting whole set of all roads is not confusing. Since this number may be too large, print the answer modulo 109 + 7.

Examples

input

3
2 3 1

output

input

4
2 1 1 1

output

input

5
2 4 2 5 3

output

Note

Consider the first sample case. There are 3 towns and 3 roads. The towns are numbered from 1 to 3 and the roads are $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ initially. Number the roads 1 to 3 in this order.

The sets of roads that ZS the Coder can flip (to make them not confusing) are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}. Note that the empty set is invalid because if no roads are flipped, then towns 1, 2, 3 is form a directed cycle, so it is confusing. Similarly, flipping all roads is confusing too. Thus, there are a total of 6 possible sets ZS the Coder can flip.

The sample image shows all possible ways of orienting the roads from the first sample such that the network is not confusing.

$\text{[math]}$

分析：可能存在路径，环，或者两者结合，只需统计每个环上的边数p，对答案的贡献是pow(2,p)-2；

　　　剩余的路的边数q对答案的贡献是pow(2，q)，然后所有的乘起来即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=2e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,cycle,co[maxn],vis[maxn],a[maxn];
ll ans;
void dfs1(int now)
{
    co[cycle]++;
    vis[now]=3;
    if(vis[a[now]]==3)return;
    dfs1(a[now]);
}
void dfs(int now)
{
    vis[now]=2;
    if(!vis[a[now]])
    {
        dfs(a[now]);
    }
    else if(vis[a[now]]==1)
    {
        vis[now]=1;
        return;
    }
    else
    {
        dfs1(now);
        cycle++;
    }
    vis[now]=1;
}
int main()
{
    int i,j;
    ans=1;
    scanf("%d",&n);
    rep(i,1,n)scanf("%d",&a[i]);
    rep(i,1,n)if(!vis[i])dfs(i);
    rep(i,0,cycle-1)ans=ans*(qpow(2,co[i])-2+mod)%mod,n-=co[i];
    printf("%lld
",ans*qpow(2,n)%mod);
    //system("Pause");
    return 0;
}