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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
using ll = long long;
using db = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
template <class T>
using PR = pair<T, T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
const int MX = 2e5 + 5;
const int MOD = 1e9 + 7;
const ll INF = 1e18;
const db PI = acos((db)-1);
const int xdir[4] = {1, 0, -1, 0}, ydir[4] = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
T fstTrue(T lo, T hi, U first) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
template <class T, class U>
T lstTrue(T lo, T hi, U first) {
  lo--;
  assert(lo <= hi);
  while (lo < hi) {
    T mid = lo + (hi - lo + 1) / 2;
    first(mid) ? lo = mid : hi = mid - 1;
  }
  return lo;
}
template <class T>
void remDup(V<T>& v) {
  sort(begin(v), end(v));
  v.erase(unique(begin(v), end(v)), end(v));
}
template <typename T1, typename T2>
str to_string(pair<T1, T2> p);
template <typename T1, typename T2, typename T3>
str to_string(tuple<T1, T2, T3> p);
template <typename T1, typename T2, typename T3, typename T4>
str to_string(tuple<T1, T2, T3, T4> p);
str to_string(const str& S) { return '"' + S + '"'; }
str to_string(const char* S) { return to_string((str)S); }
str to_string(bool b) { return b ? "true" : "false"; }
template <class T>
str to_string(complex<T> c) {
  stringstream ss;
  ss << c;
  return ss.str();
}
template <size_t SZ>
str to_string(bitset<SZ> v) {
  str res = "";
  for (int i = (0); i < (SZ); ++i) res += char(v[i] + '0');
  return res;
}
template <typename T>
str to_string(T v) {
  bool fst = 1;
  str res = "";
  for (const auto& x : v) {
    if (!fst) res += " ";
    fst = 0;
    res += to_string(x);
  }
  return res;
}
template <typename T1, typename T2>
str to_string(pair<T1, T2> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
void DBG() { cerr << endl; }
template <typename Head, typename... Tail>
void DBG(Head H, Tail... T) {
  cerr << " " << to_string(H);
  DBG(T...);
}
void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIn(str name) { freopen(name.c_str(), "r", stdin); }
void setOut(str name) { freopen(name.c_str(), "w", stdout); }
void setIO(str name = "") {
  unsyncIO();
  setPrec();
  if ((int)((name).size())) setIn(name + ".in"), setOut(name + ".out");
}
struct DSU {
  vi e;
  void init(int N) { e = vi(N, -1); }
  int get(int x) { return e[x] < 0 ? x : e[x] = get(e[x]); }
  bool unite(int x, int y) {
    if (((x = get(x)) == (y = get(y)))) return 0;
    if (e[x] > e[y]) swap(x, y);
    e[x] += e[y];
    e[y] = x;
    return 1;
  }
};
int32_t main() {
  setIO();
  int n;
  cin >> n;
  DSU D;
  D.init(n);
  for (int i = (0); i < (n); ++i) {
    int x;
    cin >> x;
    --x;
    D.unite(i, x);
  }
  set<int> res;
  for (int i = (0); i < (n); ++i) res.insert(D.get(i));
  cout << (int)((res).size()) << "\n";
}