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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;

import java.util.*;

public class A_GENERAL {
   
	// for fast output printing  : use printwriter or stringbuilder
	// remember to close pw using pw.close()
	static PrintWriter pw =new PrintWriter(new OutputStreamWriter(System.out));
	static StringBuilder sb = new StringBuilder();
	static long seive_size = (long) 1e6;
    static ArrayList<Integer> primes = new ArrayList<>();
    static boolean[] set = new boolean[(int) seive_size+1];
    static int n;
    static ArrayList<Integer>[] adj;
    static boolean[] visited;
    static ArrayDeque<Integer> q = new ArrayDeque<>();
    static final long MOD = 998244353;
    
	public static void main(String[] args) {
		MyScanner sc = new MyScanner();
		n = sc.nextInt();
		init(n);
		for(int i = 1; i <= n; i++) {
			int u = sc.nextInt();
			adj[i].add(u);
			adj[u].add(i);
		}
		int c = 0;
		for(int i = 1; i <= n; i++) {
			if(!visited[i]) {
				c++;
				dfs(i);
			}
		}
		System.out.println(c);
	}
	public static void dfs(int u) {
		visited[u] = true;
		for(int v : adj[u]) {
			if(!visited[v])
				dfs(v);
		}
	}
	
	public static class Pair {
		long f;
		long s;
		Pair(long f, long s) {
			this.f = f;
			this.s = s;
		}
	}
	public static void init(int n) {
    	adj = new ArrayList[n+1];
    	visited = new boolean[n+1];
    	for(int i = 1; i <= n; i++)
			adj[i] = new ArrayList();
    }
	// print string "s" multiple times 
	// prefer to use this function for multiple printing
	public static String mp(String s, int times) {
		return String.valueOf(new char[times]).replace("\0", s);
	}
	// take log with base 2
 	public static long log2(long k) {
 		return 63-Long.numberOfLeadingZeros(k);
 	}

 	// using lambda function for sorting
   	public static void lambdaSort(long[][] arr) {
   		Arrays.sort(arr, (a, b) -> Double.compare(a[0], b[0]));
   	}
  	
  	// (n choose k) = (n/k) * ((n-1) choose (k-1))
  	public static long choose(long n, long k) {
 		return (k == 0) ? 1 : (n*choose(n-1, k-1))/k;
 	}

 	// just for keeping gcd function for other personal purposes
 	public static long gcd(long a, long b) {
 		return (a == 0) ? b : gcd(b%a, a);
 	}
 	
	public static long max(long... as) {
        long max = Long.MIN_VALUE;
        for (long a : as) max = Math.max(a, max);
        return max;
    }

    public static long min(int... as) {
        long min = Long.MAX_VALUE;
        for (long a : as) min = Math.min(a, min);
        return min;
    }
 	
    public static long modpow(long x, long n, long mod) {
    	if(n == 0) return 1%mod;
    	long u = modpow(x, n/2, mod);
    	u = (u*u)%mod;
    	if(n%2 == 1) u = (u*x)%mod;
    	return u;
    }
 	// ======================= binary search (lower and upper bound) =======================
	public static int lowerBound(long[] a, int x) {
		int lo = 0;
 		int hi = a.length-1;
 		int ans = -1;
 		while(lo <= hi) {
 			int mid = (lo+hi)/2;
 			if(x < a[mid]) {
 				hi = mid-1;
 			} else if(x > a[mid]) {
 				lo = mid+1;
 			} else if(lo != hi) {
 				 hi = mid-1; 		// for first occurrence
 				 ans = mid;
 			} else {
 				return mid;
 			}
 		}
 		return ans;
	}
	
	public static int upperBound(long[] a, long x) {
		int lo = 0;
		int hi = a.length-1;
		int ans = -1;
		while(lo <= hi) {
			int mid = (lo+hi)/2;
			if(x < a[mid]) {
				hi = mid-1;
			} else if(x > a[mid]) {
				lo = mid+1;
			} else if(lo != hi) {
				lo = mid+1;			// for last occurrence
				ans = mid;
			} else {
				return mid;
			}
		}
		return ans;
	}
	// ================================================================

	// ================== SEIVE OF ERATOSTHENES =======================
	// Complexity : O(N * log(log(N)))       ( almost O(N) )
	public static void generatePrimes() {
		// set.add(0);
		// set.add(1);
		Arrays.fill(set, true);
		set[0] = false;
		set[1] = false;
		for(int i = 2; i <= seive_size; i++) {
			if(set[i]) {
				for(long j = (long) i*i; j <= seive_size; j+=i) 
					set[(int)j] = false;
				primes.add(i);
			}
		}
	}
  
	public static boolean isPrime(long N) {
		if(N <= seive_size) return set[(int)N];
		for (int i = 0; i < (int)primes.size(); i++)
			if (N % primes.get(i) == 0) return false;
		return true;
	}
	// ===========================================================
  
 	// ================ Permutation of String ====================
 	public static void permute(String str) {
 		permute(str, 0, str.length()-1);
 	}
 	public static void permute(String str, int l, int r) 
    { 
        if (l == r) 
            System.out.println(str); 
        else
        { 
            for (int i = l; i <= r; i++) 
            { 
                str = swap(str,l,i); 
                permute(str, l+1, r); 
                str = swap(str,l,i); 
            } 
        } 
    }
	public static String swap(String a, int i, int j) 
    { 
        char temp; 
        char[] charArray = a.toCharArray(); 
        temp = charArray[i] ; 
        charArray[i] = charArray[j]; 
        charArray[j] = temp; 
        return String.valueOf(charArray); 
    } 
	// Union-find
	void makeSet(int parent[], int rank[]) {

		int i;
		for(i = 0; i < parent.length; i++) {
			parent[i] = i;
			rank[i] = 0;
		}
	}
	public static int find(int u, int parent[]) {

		if(parent[u] == u)	return u;

		int v = find(parent[u], parent);
		parent[u] = v;
		return v;

	}
	public static boolean connected(int u, int v, int[] parent) {
		return find(u, parent) == find(v, parent);
	}
	public static void Union(int u, int v, int rank[], int parent[]) {

		int x = find(u, parent);				//root of u
		int y = find(v, parent);				//root of v

		if(x == y)		return;
		if(rank[x] == rank[y]) {
			parent[y] = x;
			rank[x]++;
		}
		else if(rank[x] > rank[y]) {
			parent[y] = x;
		}
		else {
			parent[x] = y;
		}

	}
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
		int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        int[] nextIntArray(int n, int delta) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt() + delta;
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
	}

}