D. Directed Roads

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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

6

input

4

2 1 1 1

output

8

input

5

2 4 2 5 3

output

28

题意：有N个点，每个点都向其他一个点连一条有向边，形成一个N个点，N条有向边的图， 图可能有环。现在选取其中的一些边改变方向，使得图中没有环，求改变的方法数。

思路：可以想到把图按照环分块，把在同一个环中的点染成一个颜色，把其余点，也就是说不在环中的点归为另一类。

假设环中有k条边，那么每个环中有挑选一条，两条……k - 1条边进行改变方向。(1k)+(2k)+……+(k−1k) 种， 为2k−2；剩下不在环中的边不论任意地变化都不会改变图环的数量，假设剩下有N- kk条边，则为2N−kk。乘法原理相乘，取模。每次dfs一个没有跑过的点，就行了。

复杂度O(N)

代码：

/*****************************************************/

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <map>

#include <set>

#include <ctime>

#include <stack>

#include <queue>

#include <cmath>

#include <string>

#include <vector>

#include <cstdio>

#include <cctype>

#include <cstring>

#include <sstream>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)

#define   sigma_size    26

#define   lson          l,m,v<<1

#define   rson          m+1,r,v<<1|1

#define   slch          v<<1

#define   srch          v<<1|1

#define   sgetmid       int m = (l+r)>>1

#define   LL            long long

#define   ull           unsigned long long

#define   mem(x,v)      memset(x,v,sizeof(x))

#define   lowbit(x)     (x&-x)

#define   bits(a)       __builtin_popcount(a)

#define   mk            make_pair

#define   pb            push_back

#define   fi            first

#define   se            second

const int    INF    = 0x3f3f3f3f;

const LL     INFF   = 1e18;

const double pi     = acos(-1.0);

const double inf    = 1e18;

const double eps    = 1e-9;

const LL     mod    = 1e9+7;

const int    maxmat = 10;

const ull    BASE   = 31;

/*****************************************************/

const int maxn = 2e5 + 5;

std::vector<int> G[maxn];

int pre[maxn], dfs_clock, block[maxn];

LL qpow(LL a, LL b) {

    LL res = 1;

    while (b) {

        if (b & 1) res = res * a % mod;

        a = a * a % mod;

        b >>= 1;

    }

    return res;

}

int dfs(int u, int c) {

    block[u] = c;

    if (pre[u]) {

        int tmp = dfs_clock + 1 - pre[u];

        pre[u] = ++dfs_clock;

        return tmp;

    }

    else {

        pre[u] = ++dfs_clock;

        for (int i = 0; i < G[u].size(); i ++) {

            int v = G[u][i];

            if (block[v] && block[v] != c) continue;

            return dfs(v, c);

        }

        return 0;

    }

}

void work(int N) {

    mem(pre, 0);

    mem(block, 0);

    dfs_clock = 0;

    int ans = 0, color = 0;

    LL res = 1;

    for (int i = 1; i <= N; i ++) {

        if (!pre[i]) {

            int loop = dfs(i, ++color);

            ans += loop;

            if (loop) res = res * (qpow(2, loop) - 2) % mod;

        }

    }

    res = res * qpow(2, N - ans) % mod;

    cout<<res<<endl;

}

int main(int argc, char const *argv[]) {

    int N;

    cin>>N;

    for (int i = 1; i <= N; i ++) {

        int k;

        scanf("%d", &k);

        G[i].pb(k);

    }

    work(N);

    return 0;

}

萌新第一次发博客，写得不好见谅。