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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, const char * argv[]) {
   
    
    int N;
    scanf("%d\n",&N);
    
    int *a;
    
    a=(int *)malloc(N+1);
  
    char *buff;
    
    buff=(char *)malloc(N*2+1);
    
    gets(buff);
    
    char str2[256] = " ";
    strcat(buff,str2);
    int nn=1;
    for (int i=1; i<=N*2+1; i++)
    {
        if (buff[i]==' ')
        {
            a[nn]=atoi(&buff[i-1]);
            nn++;

        }
        
        
    }
    
    int likesum=0;
    
    for (int i=1; i<=N; i++) {
        
        // printf("  %d    %d \n",a[i],a[a[i]]);
        
        if (i==a[a[i]]) {
            likesum++;
            //printf("\na");
        }
        
        
    }
    
    
    printf("%d\n",likesum/2);
    
    
    return 0;
}