// Solution to "Movie Night". Makes use of formula at:
// https://cs.stackexchange.com/questions/123381/number-of-connected-induced-subgraph-in-a-tree

import java.util.*;
import java.io.*;

public class RSRMovieNight {
  public static long A[],B[];
  public static int S[];
  public static boolean vis[];
  public static ArrayList par[];
  public static final long BIG = 1000000007;

  public static void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(in.readLine().strip());
    A = new long[n];
    B = new long[n];
    Arrays.fill(B,1L);
    S = new int[n];
    vis = new boolean[n];
    Arrays.fill(vis,false);
    par = new ArrayList[n];
    for (int i = 0; i < n; i++) par[i] = new ArrayList<Integer>();

    for (int i = 0; i < n; i++) {
      int x = Integer.parseInt(in.readLine().strip());
      par[x-1].add(i);
    }

    long ans = 1L;
    int search = 0;
    for (int i = 0; i < n; i++) {
      if (vis[i]) continue;
      search += 1;
      long ret = dfs(i,search);
      if (ret > 0) ans = (ans*(1L+ret))%BIG;
    }
    System.out.println((ans-1L+BIG)%BIG);
  }


  public static long dfs(int i, int search) {
    vis[i] = true;
    S[i] = search;    
    boolean loop = false;
    long mult = 0;
    long result = 0;
    for (Object j1 : par[i]) {
      int j = (int)j1;
      if (!vis[j]) {
        result = dfs(j,search);
        if (result > 0) { // backing up from a loop node
          mult = result;
        } else {
          A[i] = (A[i]+A[j]+B[j])%BIG;
          B[i] = (B[i]*(1+B[j]))%BIG;
        }
      } else if (S[j] != search) {
        A[i] = (A[i]+A[j]+B[j])%BIG;
        B[i] = (B[i]*(1+B[j]))%BIG;
      } else { // we're the last node in the loop--this can happen only once
        loop = true;
      }
    }
    if (loop) {
      return B[i];
    } else {
      return (B[i]*mult)%BIG; // mult will be zero for non-loop nodes
    }
  }
}