Problem:
Score : 400 points

Problem Statement

We have a permutation of the integers from 1 through N, p_1, p_2, .., p_N.
We also have M pairs of two integers between 1 and N (inclusive), represented as (x_1,y_1), (x_2,y_2), .., (x_M,y_M).
AtCoDeer the deer is going to perform the following operation on p as many times as desired so that the number of i (1 ≤ i ≤ N) such that p_i = i is maximized:

Choose j such that 1 ≤ j ≤ M, and swap p_{x_j} and p_{y_j}.

Find the maximum possible number of i such that p_i = i after operations.

Constraints

2 ≤ N ≤ 10^5

1 ≤ M ≤ 10^5

p is a permutation of integers from 1 through N.

1 ≤ x_j,y_j ≤ N

x_j ≠ y_j

If i ≠ j, \{x_i,y_i\} ≠ \{x_j,y_j\}.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
p_1 p_2 .. p_N
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the maximum possible number of i such that p_i = i after operations.

Sample Input 1

5 2
5 3 1 4 2
1 3
5 4

Sample Output 1

2

If we perform the operation by choosing j=1, p becomes 1 3 5 4 2, which is optimal, so the answer is 2.

Sample Input 2

3 2
3 2 1
1 2
2 3

Sample Output 2

3

If we perform the operation by, for example, choosing j=1, j=2, j=1 in this order, p becomes 1 2 3, which is obviously optimal.
Note that we may choose the same j any number of times.

Sample Input 3

10 8
5 3 6 8 7 10 9 1 2 4
3 1
4 1
5 9
2 5
6 5
3 5
8 9
7 9

Sample Output 3

8

Sample Input 4

5 1
1 2 3 4 5
1 5

Sample Output 4

5

We do not have to perform the operation.

Solution in C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

class Program
{
    static void Main(string[] args)
    {
        new Program().Calc();
    }

    Scanner cin;

    public Program() { }

    void Calc()
    {
        cin = new Scanner();

        int n = cin.nextInt();
        int m = cin.nextInt();

        int[] p = cin.ArrayInt(n);

        int[] x = new int[m];
        int[] y = new int[m];

        var unionFind = new UnionFindTree(n);

        for (int i = 0; i < m; ++i)
        {
            x[i] = cin.nextInt();
            y[i] = cin.nextInt();

            unionFind.Unite(x[i], y[i]);
        }

        int count = 0;

        for (int i = 0; i < n; ++i)
        {
            if (i+1 != p[i])
            {
                if (unionFind.Same(i + 1, p[i])) { ++count; }
            }
            else
            {
                ++count;
            }
        }

        Console.WriteLine(count);
    }
}

class UnionFindTree
{
    int[] parent;

    public UnionFindTree(int n)
    {
        parent = new int[n+1];

        for (int i = 0; i <= n; ++i)
        {
            parent[i] = i;
        }
    }

    public int Root(int x)
    {
        int root = x;
        while(parent[root] != root)
        {
            root = parent[root];
        }

        return root;
    }

    public void Unite(int x, int y)
    {
        int rootX = Root(x);
        int rootY = Root(y);
        if (rootX != rootY)
        {
            parent[rootX] = rootY;
        }
    }

    public bool Same(int x, int y)
    {
        return Root(x) == Root(y);
    }
}

class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}