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;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const long long linf = 4000000000000000000LL;
const long long inf = 998244353;
const long double pi = 3.1415926535;
long long maximum(long long x, long long y) {
  if (x > y) return x;
  return y;
}
long long minimum(long long x, long long y) {
  if (x < y) return x;
  return y;
}
void pv(vector<int> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<long long> a) {
  for (auto& x : a) cout << x << " ";
  cout << '\n';
}
void pv(vector<vector<int> > a) {
  for (int i = (0); i < (a.size()); ++i) {
    cout << i << '\n';
    pv(a[i]);
    cout << '\n';
  }
}
void pv(vector<vector<long long> > a) {
  for (int i = (0); i < (a.size()); ++i) {
    cout << i << '\n';
    pv(a[i]);
  }
  cout << '\n';
}
void pv(vector<string> a) {
  for (auto& x : a) cout << x << '\n';
  cout << '\n';
}
int modInverse(int a, int m) {
  int m0 = m;
  int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    int q = a / m;
    int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long power(long long x, long long y) {
  long long k = 1 << 30;
  long long z = 1;
  while (k != 0) {
    z *= z;
    z %= inf;
    if (y >= k) {
      z *= x;
      z %= inf;
      y -= k;
    }
    k >>= 1;
  }
  return z;
}
long double area(pair<int, int> x, pair<int, int> y, pair<int, int> z) {
  return ((long long)x.second * y.first + (long long)y.second * z.first +
          (long long)z.second * x.first - (long long)x.first * y.second -
          (long long)y.first * z.second - (long long)z.first * x.second) /
         2.0;
}
bool clockwise(pair<int, int> x, pair<int, int> y, pair<int, int> z) {
  return area(x, y, z) > 0;
}
struct point {
  long double x, y;
};
long double area(point x, point y, point z) {
  return (x.y * y.x + y.y * z.x + z.y * x.x - x.x * y.y - y.x * z.y -
          z.x * x.y) /
         2.0;
}
bool clockwise(point x, point y, point z) { return area(x, y, z) > 0; }
int gcd(int a, int b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(a, b % a);
}
int popcount(int a) {
  int count = 0;
  while (a != 0) {
    count += (a % 2);
    a >>= 1;
  }
  return count;
}
long long choose(long long n, long long r) {
  long long p = 1, k = 1;
  if (n - r < r) r = n - r;
  if (r != 0) {
    while (r) {
      p *= n;
      k *= r;
      long long m = gcd(p, k);
      p /= m;
      k /= m;
      n--;
      r--;
    }
  } else
    p = 1;
  return p;
}
vector<long long> prefix_hash(string& a, vector<long long>& powers) {
  int n = int(a.size());
  vector<long long> prefix(n + 1);
  for (int i = (0); i < (n); ++i)
    prefix[i + 1] = (prefix[i] + powers[i] * (a[i] - 'a' + 1)) % inf;
  return prefix;
}
void setIO(string second) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int n;
vector<vector<int> > adj;
vector<int> v;
void dfs(int i, int p) {
  v[i] = 1;
  for (auto& x : adj[i]) {
    if (x != p) dfs(x, i);
  }
}
int main() {
  setIO("test3");
  cin >> n;
  adj.resize(n);
  for (int i = (0); i < (n); ++i) {
    int a;
    cin >> a;
    adj[--a].push_back(i);
    adj[i].push_back(a);
  }
  int ans = 0;
  v.resize(n);
  for (int i = (0); i < (n); ++i) {
    if (!v[i]) {
      ans++;
      dfs(i, -1);
    }
  }
  cout << ans << '\n';
}