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;
using System.Linq;
class Program
{
    static ArrayList s = new ArrayList();
    public static ArrayList DepthFirstSearch(Node root)
    {
        { s.Add(root.x); } //rootに対する処理内容
        root.Visited = true; //rootを訪問済みにする
        if (root.Children != null) //子が存在するならば
        {
            for (int i = 0; i < root.Children.Count; i++)
            {
                if (!((Node)(root.Children[i])).Visited) //i番目の子が訪問済みでないならば
                {
                    DepthFirstSearch(((Node)root.Children[i]));
                }
            }
        }
        return s;
    }
    static void Main()
    {
        //入力を受け取る
        int[] nm = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
        int n = nm[0], m = nm[1];
        int[] p = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
        int[][] xy = new int[m][];
        for (int i = 0; i < m; i++)
            xy[i] = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();

        //処理内容
        Node[] v = new Node[n];
        for (int i = 0; i < n; i++)
            v[i] = new Node(i + 1);
        for (int i = 0; i < m; i++)
        {
            v[xy[i][0] - 1].Children.Add(v[xy[i][1] - 1]);
            v[xy[i][1] - 1].Children.Add(v[xy[i][0] - 1]);
        }
        ArrayList t = new ArrayList();
        for (int i = 0; i < n; i++)
        {
            if (!v[i].Visited)
            {
                t.Add(DepthFirstSearch(v[i]));
                s = new ArrayList();
            }
        }
        int ans = 0;
        for (int i = 0; i < t.Count; i++)
        {
            int[] nums = (int[])((ArrayList)t[i]).ToArray(typeof(int));
            int[] nums2 = new int[nums.Length];
            int l = 0;
            foreach (var j in nums)
            {
                nums2[l] = p[j - 1];
                l++;
            }
            foreach (var j in nums)
            {
                if (nums2.Count(q => q == j) > 0)
                    ans++;
            }
        }
        Console.WriteLine(ans);
    }
}
class Node //ノード
{
    public bool Visited { get; set; } //訪問済みか否か
    public ArrayList Children = new ArrayList(); //子の配列
    public int x;
    public Node(int x)
    {
        this.x = x;
    }
}