Problem:
Score : 200 points

Problem Statement

There are N rabbits, numbered 1 through N.

The i-th (1≤i≤N) rabbit likes rabbit a_i.
Note that no rabbit can like itself, that is, a_i≠i.

For a pair of rabbits i and j (i＜j), we call the pair (i，j) a friendly pair if the following condition is met.

Rabbit i likes rabbit j and rabbit j likes rabbit i.

Calculate the number of the friendly pairs.

Constraints

2≤N≤10^5

1≤a_i≤N

a_i≠i

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the number of the friendly pairs.

Sample Input 1

4
2 1 4 3

Sample Output 1

2

There are two friendly pairs: (1，2) and (3，4).

Sample Input 2

3
2 3 1

Sample Output 2

0

There are no friendly pairs.

Sample Input 3

5
5 5 5 5 1

Sample Output 3

1

There is one friendly pair: (1，5).

Solution in Scala (2.11.7) (judge verdict: Runtime Error)

object Main extends App {
    val n = io.StdIn.readInt
    val as = io.StdIn.readLine.split(" ").map(_.toInt)

    val targets = rec1(1, Map.empty)
    def rec1(index: Int, map: Map[Int, Int]): Map[Int, Int] = {
        if (n < index) map
        else {
            rec1(index+1, map+(index -> as(index-1)))
        }
    }

    val result = rec2(0, targets)
    def rec2(count: Int, map: Map[Int, Int]): Int = {
        if (map.isEmpty || map.size==1) count
        else {
            val rabbit1 = map.head
            val rabbit1_key = rabbit1._1
            val rabbit1_like = rabbit1._2
            val rabbit1_like_like = map(rabbit1_like)

            if (rabbit1_key == rabbit1_like_like) rec2(count+1, map-rabbit1_key-rabbit1_like)
            else rec2(count, map-rabbit1_key-rabbit1_like)

        }
    }
    
    println(result)
}