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: Time Limit Exceeded)

#include <algorithm>
#include <array>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>

#define REP(X, Y, Z) for (ll(X) = (Y); (X) < (Z); (X)++)
#define REV(X, Y, Z) for (ll(X) = (Y); (X) > (Z); (X)--)

using namespace std;
using ll = long long;
using dll = deque<ll>;
using pll = pair<ll, ll>;
using qll = queue<ll>;
using usll = unordered_set<ll>;
using umll = unordered_map<ll, ll>;
using sll = stack<int>;
using vll = vector<ll>;
using vvll = vector<vll>;

constexpr ll kMod = 1e9 + 7;

ll gcd(ll a, ll b) {
  if (a < b) {
    return gcd(b, a);
  }
  ll c = a % b;
  while (c != 0) {
    a = b;
    b = c;
    c = a % b;
  }
  return b;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

void init() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}

int main() {
  init();

  ll N;
  cin >> N;
  vll A(N);
  REP(i, 0, N) {
    ll a;
    cin >> a;
    A[i] = a - 1;
  }

  ll ans = 0;
  REP(i, 0, N - 1) {
    REP(j, i, N) {
      if (i == A[j] && A[i] == j) {
        ans++;
      }
    }
  }
  cout << ans << endl;

  return 0;
}