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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, s, e) for (int i = s; i < e; ++i)
#define sort_(a) stable_sort(a.begin(), a.end())
#define rsort(a) stable_sort(a.rbegin(), a.rend())
#define sum(a) accumulate(a.begin(), a.end(), 0LL)
#define join(a, d) accumulate(a.begin() + 1, a.end(), a[0], [](string s, string t) {return s + d + t;})
#define all(a) a.begin(), a.end()
#define __lcm(a, b) std::__detail::__lcm(a, b)
typedef long long ll;
const long mod = 1e9 + 7;

int main(void) {
#ifdef DEBUG
  freopen("input.txt", "r", stdin);
#endif

  ios_base::sync_with_stdio(false);
  cin.tie(0);

  int N;
  cin >> N;
  vector<int> A(N + 1, 0);
  rep(i, 1, N + 1) {
    cin >> A[i];
  }
  pair<int, int> p;
  map<pair<int, int>, long> cnt{};
  rep(i, 1, N + 1) {
    p.first = min(i, A[i]);
    p.second = max(i, A[i]);
    cnt[p] += 1;
  }
  long ans = 0;
  for (auto n : cnt) {
    if (n.second == 2)
      ans += 1;
  }
  cout << ans << endl;
  return 0;
}