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.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C391vc {
	static int MOD=(int)Math.pow(10,9)+7;
	static int root[];
	static int size[];
	public static void main(String args[]) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		StringBuilder sb = new StringBuilder();
		// ----------My Code----------
		
		int n=in.nextInt();
		root=new int[n];
		size=new int[n];
		Arrays.fill(size,1);
		
		for(int i=0;i<n;i++){
			root[i]=i;
		}
		for(int i=0;i<n;i++){
			modified_Union(i,in.nextInt()-1);
		}
		
		int cnt=0;
		for(int i=0;i<n;i++){
			if(root[i]==i)
				cnt++;
		}
		System.out.println(cnt);
		// ---------------The End------------------

		out.close();
	}

	public static int modified_root(int i){
		while(root[i]!=i){
			root[i]=root[root[i]];
			i=root[i];	
		}
		
		return i;
	}
	
	public static void modified_Union(int a, int b){
		
		int roota=modified_root(a);
		int rootb=modified_root(b);
		
		if(roota == rootb) 
			return;
		
		if(size[roota]>size[rootb]){
			root[rootb]=root[roota];
			size[roota]+=size[rootb];
			
		}
		else{
			root[roota]=root[rootb];
			size[rootb]+=size[roota];
			
		}
		
	}
	
	public static boolean find(int a,int b){

		int roota=modified_root(a);
		int rootb=modified_root(b);
		
		if(roota==rootb)
			return true;
		return false;
		
	}
	// ---------------Extra Methods------------------

	public static long pow(long x, long n, long mod) {
		long res = 1;
		x %= mod;
		while (n > 0) {
			if (n % 2 == 1) {
				res = (res * x) % mod;
			}
			x = (x * x) % mod;
			n /= 2;
		}
		return res;
	}

	public static boolean isPal(String s) {
		for (int i = 0, j = s.length() - 1; i <= j; i++, j--) {
			if (s.charAt(i) != s.charAt(j))
				return false;
		}
		return true;
	}

	public static String rev(String s) {
		StringBuilder sb = new StringBuilder(s);
		sb.reverse();
		return sb.toString();
	}

	public static long gcd(long x, long y) {
		if (x % y == 0)
			return y;
		else
			return gcd(y, x % y);
	}

	public static int gcd(int x, int y) {
		if (x % y == 0)
			return y;
		else
			return gcd(y, x % y);
	}

	public static long gcdExtended(long a, long b, long[] x) {

		if (a == 0) {
			x[0] = 0;
			x[1] = 1;
			return b;
		}
		long[] y = new long[2];
		long gcd = gcdExtended(b % a, a, y);

		x[0] = y[1] - (b / a) * y[0];
		x[1] = y[0];

		return gcd;
	}

	public static int abs(int a, int b) {
		return (int) Math.abs(a - b);
	}

	public static long abs(long a, long b) {
		return (long) Math.abs(a - b);
	}

	public static int max(int a, int b) {
		if (a > b)
			return a;
		else
			return b;
	}

	public static int min(int a, int b) {
		if (a > b)
			return b;
		else
			return a;
	}

	public static long max(long a, long b) {
		if (a > b)
			return a;
		else
			return b;
	}

	public static long min(long a, long b) {
		if (a > b)
			return b;
		else
			return a;
	}
	// ---------------Extra Methods------------------

	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine() {
			String fullLine = null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine = reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public char nextChar() {
			return next().charAt(0);
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n, int f) {
			if (f == 0) {
				int[] arr = new int[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextInt();
				}
				return arr;
			} else {
				int[] arr = new int[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextInt();
				}
				return arr;
			}
		}

		public long[] nextLongArray(int n, int f) {
			if (f == 0) {
				long[] arr = new long[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextLong();
				}
				return arr;
			} else {
				long[] arr = new long[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextLong();
				}
				return arr;
			}

		}

		public double[] nextDoubleArray(int n, int f) {
			if (f == 0) {
				double[] arr = new double[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			} else {
				double[] arr = new double[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			}
		}

	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			if (this.a != o.a)
				return Integer.compare(this.a, o.a);
			else
				return Integer.compare(this.b, o.b);
		}

		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.a == a && p.b == b;
			}
			return false;
		}

		public int hashCode() {
			return new Integer(a).hashCode() * 31 + new Integer(b).hashCode();
		}

	}

}