Codeforces 711D Directed Roads【Dfs+思维】

D. Directed Roads

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

ZS the Coder and Chris the Baboon has explored Udayland for quite some time. They realize that it consists of n towns numbered from 1 to n.

There are n directed roads in the Udayland. i-th of them goes from town i to some other town a_i (a_i ≠ 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, ..., A_k (k > 1) such that for every 1 ≤ i < k there is a road from town A_i to town A_i + 1 and another road from town A_k 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, ..., a_n (1 ≤ a_i ≤ n, a_i ≠ i), a_i denotes a road going from town i to town a_i.

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

6

Input

4
2 1 1 1

Output

8

Input

5
2 4 2 5 3

Output

28

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 , , 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.

题目大意：

给你一个有向图，其中包含n个点以及n条边，我们可以将任意边反向，问当前图中有多少种方案数使得图中没有有向环。

思路：

1、首先我们Dfs整个图找环，并且统计出每个环中包含的点的个数（同时也就是包含的边的个数）。对应当前这个环来讲，不考虑其他部分，只考虑这个环的话，其可行方案（没有环）有：2^（环中点的个数）-2；其中减去的一个是当前有向环，另一个是将所有边反向的环。

2、那么我们对应将所有环的可行方案累乘，得到这个图中有环部分的可行方案，记做output。

3、那么我们再记录一下不在环中的点的个数，再令output*=2^（不在环中的点的个数）；那么此时output就是最终答案。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
#define ll __int64
#define mod 1000000007
vector<int >mp[200040];
int vis[200040];
int dfn[200040];
int num[200040];
ll output,sum,n,sig,cnt;
ll poww(ll a,ll b)
{
    ll ans=1;
    a%=mod;
    while(b>0)
    {
        if(b%2==1)ans=(ans*a)%mod;
        b/=2;
        a=(a*a)%mod;
    }
    return ans;
}
void Dfs(int u)
{
    num[u]=cnt++;
    vis[u]=1;
    dfn[u]=sig;
    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i];
        if(vis[v]==1&&dfn[v]==sig)
        {
            ll tot=num[u]-num[v]+1;
            output=(output*((poww(2,tot)-2)))%mod;
            sum+=tot;
        }
        if(vis[v]==0)
        Dfs(v);
    }
}
int main()
{
    while(~scanf("%I64d",&n))
    {
        for(int i=1;i<=n;i++)mp[i].clear();
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            mp[i].push_back(x);
        }
        memset(dfn,0,sizeof(dfn));
        memset(vis,0,sizeof(vis));
        output=1;
        cnt=1;
        sig=0;
        sum=0;
        for(int i=1;i<=n;i++)
        {
            if(vis[i]==0)
            {
                sig++;
                Dfs(i);
            }
        }
        output=(output*(poww(2,n-sum)))%mod;
        printf("%I64d\n",output%mod);
    }
}