Problem: PolandBall lives in a forest with his family. There are some trees in the forest. Trees are undirected acyclic graphs with k vertices and k - 1 edges, where k is some integer. Note that one vertex is a valid tree.

There is exactly one relative living in each vertex of each tree, they have unique ids from 1 to n. For each Ball i we know the id of its most distant relative living on the same tree. If there are several such vertices, we only know the value of the one with smallest id among those.

How many trees are there in the forest?

Input

The first line contains single integer n (1 ≤ n ≤ 104) — the number of Balls living in the forest.

The second line contains a sequence p1, p2, ..., pn of length n, where (1 ≤ pi ≤ n) holds and pi denotes the most distant from Ball i relative living on the same tree. If there are several most distant relatives living on the same tree, pi is the id of one with the smallest id.

It's guaranteed that the sequence p corresponds to some valid forest.

Hacking: To hack someone, you should provide a correct forest as a test. The sequence p will be calculated according to the forest and given to the solution you try to hack as input. Use the following format:

In the first line, output the integer n (1 ≤ n ≤ 104) — the number of Balls and the integer m (0 ≤ m < n) — the total number of edges in the forest. Then m lines should follow. The i-th of them should contain two integers ai and bi and represent an edge between vertices in which relatives ai and bi live. For example, the first sample is written as follows:
    
    
      
    5 3  
    1 2  
    3 4  
    4 5  
    

Output

You should output the number of trees in the forest where PolandBall lives.

Interaction

From the technical side, this problem is interactive. However, it should not affect you (except hacking) since there is no interaction.

Examples

Input

5
2 1 5 3 3

Output

2

Input

1
1


Output

1

Note

In the first sample testcase, possible forest is: 1-2 3-4-5. 

There are 2 trees overall.

In the second sample testcase, the only possible graph is one vertex and no edges. Therefore, there is only one tree.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int xd[] = {1, 0, -1, 0};
int yd[] = {0, -1, 0, 1};
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
inline bool isEqual(double x, double y) {
  const double epsilon = 1e-5;
  return abs(x - y) <= epsilon * abs(x);
}
long double triarea(long long x1, long long y1, long long x2, long long y2,
                    long long x3, long long y3) {
  return (long double)abs(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2)) /
         2.0;
}
bool triinside(long x1, long y1, long x2, long y2, long x3, long y3, long x,
               long y) {
  long area1 = triarea(x1, y1, x2, y2, x3, y3);
  long area2 = triarea(x, y, x2, y2, x3, y3);
  long area3 = triarea(x1, y1, x, y, x3, y3);
  long area4 = triarea(x1, y1, x2, y2, x, y);
  if (isEqual(area1, area2 + area3 + area4)) return true;
  return false;
}
long long dist(long long x1, long long y1, long long x2, long long y2) {
  return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
}
const double pi = acos(-1.0);
long double todegree(long double radian) {
  if (radian < 0) radian += 2 * pi;
  return radian * 180 / pi;
}
long double toradian(long double degree) { return degree * pi / 180.0; }
long long poww(long long a, int b) {
  long long ret = 1;
  for (int i = 0; i < b; i++) ret *= a;
  return ret;
}
long long sumOfNumbers(long long num) {
  long long ret = 0;
  while (num) {
    ret += num % 10;
    num /= 10;
  }
  return ret;
}
unsigned long long fact(unsigned long long n) {
  unsigned long long sum = 1;
  while (n > 1) {
    sum *= n;
    n--;
  }
  return sum;
}
unsigned long long combination(unsigned long long a, unsigned long long b) {
  long long ret = 1;
  for (int i = 0; i < b; i++) ret = ret * (a - i) / (i + 1);
  return ret;
}
bool v[100005];
vector<vector<int>> g(100005);
vector<int> vo;
void dfs(int node) {
  v[node] = true;
  for (auto child : g[node]) {
    if (!v[child]) dfs(child);
  }
}
int main() {
  std::ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) {
    int o;
    cin >> o;
    g[i + 1].push_back(o);
    g[o].push_back(i + 1);
    vo.push_back(o);
  }
  int ans = 0;
  for (int i = 1; i <= (int)(n); ++i) {
    if (!v[vo[i - 1]]) {
      dfs(vo[i - 1]);
      ans++;
    }
  }
  cout << ans;
  return 0;
}