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.util.*;


public class p_757c
{
    private int[] parent; // parent[i] = parent of i

    private byte[] rank; // rank[i] = rank of subtree rooted at i (never more
                         // than 31)

    private int count; // number of components


    /**
     * Initializes an empty union�find data structure with {@code n} sites
     * {@code 0} through {@code n-1}. Each site is initially in its own
     * component.
     *
     * @param n
     *            the number of sites
     * @throws IllegalArgumentException
     *             if {@code n < 0}
     */
    public p_757c( int n )
    {
        if ( n < 0 )
            throw new IllegalArgumentException();
        count = n;
        parent = new int[n];
        rank = new byte[n];
        for ( int i = 0; i < n; i++ )
        {
            parent[i] = i;
            rank[i] = 0;
        }
    }


    /**
     * Returns the component identifier for the component containing site
     * {@code p}.
     *
     * @param p
     *            the integer representing one site
     * @return the component identifier for the component containing site
     *         {@code p}
     * @throws IndexOutOfBoundsException
     *             unless {@code 0 <= p < n}
     */
    public int find( int p )
    {
        validate( p );
        while ( p != parent[p] )
        {
            parent[p] = parent[parent[p]]; // path compression by halving
            p = parent[p];
        }
        return p;
    }


    /**
     * Returns the number of components.
     *
     * @return the number of components (between {@code 1} and {@code n})
     */
    public int count()
    {
        return count;
    }


    /**
     * Returns true if the the two sites are in the same component.
     *
     * @param p
     *            the integer representing one site
     * @param q
     *            the integer representing the other site
     * @return {@code true} if the two sites {@code p} and {@code q} are in the
     *         same component; {@code false} otherwise
     * @throws IndexOutOfBoundsException
     *             unless both {@code 0 <= p < n} and {@code 0 <= q < n}
     */
    public boolean connected( int p, int q )
    {
        return find( p ) == find( q );
    }


    /**
     * Merges the component containing site {@code p} with the the component
     * containing site {@code q}.
     *
     * @param p
     *            the integer representing one site
     * @param q
     *            the integer representing the other site
     * @throws IndexOutOfBoundsException
     *             unless both {@code 0 <= p < n} and {@code 0 <= q < n}
     */
    public void union( int p, int q )
    {
        int rootP = find( p );
        int rootQ = find( q );
        if ( rootP == rootQ )
            return;

        // make root of smaller rank point to root of larger rank
        if ( rank[rootP] < rank[rootQ] )
            parent[rootP] = rootQ;
        else if ( rank[rootP] > rank[rootQ] )
            parent[rootQ] = rootP;
        else
        {
            parent[rootQ] = rootP;
            rank[rootP]++;
        }
        count--;
    }


    // validate that p is a valid index
    private void validate( int p )
    {
        int n = parent.length;
        if ( p < 0 || p >= n )
        {
            throw new IndexOutOfBoundsException( "index " + p + " is not between 0 and " + ( n - 1 ) );
        }
    }


    public static void main( String[] args )
    {
        Scanner stdin = new Scanner( System.in );

        int n = stdin.nextInt();

        p_757c uf = new p_757c( n );
        for ( int i = 0; i < n; i++ )
        {
            int p = i;
            int q = stdin.nextInt() - 1;

            if ( uf.connected( p, q ) )
                continue;
            uf.union( p, q );

        }

        System.out.println( uf.count() );

    }
}