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: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
using E = System.Linq.Enumerable;

class Program
{
    static void Main(string[] args)
    {
        //入力を受け取る
        var NM = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
        var N = NM[0];
        var M = NM[1];
        
        var P = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
        var UF = new UnionFind(N+1);
      
        for(int i = 0;i < M;i++){
          var XY = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
          var X = XY[0];
          var Y = XY[1];
          UF.Union(X,Y);
        }
        var ans = 0;
        for(int i = 0;i < N;i++){
          if(UF.Same(P[i],i+1)){
            ans++;
          }
        }
        Console.WriteLine(ans);
    }
  
class UnionFind
{
    // 親要素のインデックスを保持する
    // 親要素が存在しない(自身がルートである)とき、マイナスでグループの要素数を持つ
    public int[] Parents { get; set; }
    public UnionFind(int n)
    {
        this.Parents = new int[n];
        for (int i = 0; i < n; i++)
        {
            // 初期状態ではそれぞれが別のグループ(ルートは自分自身)
            // ルートなのでマイナスで要素数(1個)を保持する
            this.Parents[i] = -1;
        }
    }

    // 要素xのルート要素はどれか
    public int Find(int x)
    {
        // 親がマイナスの場合は自分自身がルート
        if (this.Parents[x] < 0) return x;
        // ルートが見つかるまで再帰的に探す
        // 見つかったルートにつなぎかえる
        this.Parents[x] = Find(this.Parents[x]);
        return this.Parents[x];
    }

    // 要素xの属するグループの要素数を取得する
    public int Size(int x)
    {
        // ルート要素を取得して、サイズを取得して返す
        return -this.Parents[this.Find(x)];
    }

    // 要素x, yが同じグループかどうか判定する
    public bool Same(int x, int y)
    {
        return this.Find(x) == this.Find(y);
    }

    // 要素x, yが属するグループを同じグループにまとめる
    public bool Union(int x, int y)
    {
        // x, y のルート
        x = this.Find(x);
        y = this.Find(y);
        // すでに同じグループの場合処理しない
        if (x == y) return false;

        // 要素数が少ないグループを多いほうに書き換えたい
        if (this.Size(x) < this.Size(y))
        {
            var tmp = x;
            x = y;
            y = tmp;
        }
        // まとめる先のグループの要素数を更新
        this.Parents[x] += this.Parents[y];
        // まとめられるグループのルートの親を書き換え
        this.Parents[y] = x;
        return true;
    }
}  


}