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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;

namespace AtCoder.D
{
    public class Program
    {
        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            var (N, M) = Read2Longs();
            // インデックス変換
            var p = ReadLongs().Select(x => x - 1).ToList();
            var x = new long[M];
            var y = new long[M];
            var uft = new UnionFindTree(N);
            for (int i = 0; i < M; i++)
            {
                (x[i], y[i]) = Read2Longs();
                // インデックス変換
                x[i]--; y[i]--;
                uft.Union(x[i], y[i]);
            }

            var max = 0;
            for (int i = 0; i < N; i++)
            {
                if (i == p[i])
                {
                    max++;
                }
                else if(uft.IsUnited(i, p[i]))
                {
                    max++;
                }
            }

            return max;
        }

        #region Console.Read

        public static string ReadText() { return Console.ReadLine(); }
        public static List<string> ReadTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long ReadLong() { return long.Parse(Console.ReadLine()); }
        public static (long, long) Read2Longs() { var ls = ReadLongs(); return (ls[0], ls[1]); }
        public static (long, long, long) Read3Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2]); }
        public static (long, long, long, long) Read4Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2], ls[3]); }
        public static (long, long, long, long, long) Read5Longs() { var ls = ReadLongs(); return (ls[0], ls[1], ls[2], ls[3], ls[4]); }
        public static List<long> ReadLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }

        #endregion
    }
    public class UnionFindTree
    {
        /// <summary>
        /// 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
        /// </summary>
        public long[] Parents { get; private set; }

        public UnionFindTree(long n)
        {
            Parents = new long[n];
            for (var 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
                Parents[i] = -1;
            }
        }

        public bool Union(long x, long y)
        {
            var xRoot = Find(x);
            var yRoot = Find(y);

            // すでに同じグループの場合処理しない
            if (xRoot == yRoot) 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 (GetSize(xRoot) < GetSize(yRoot))
            {
                var tmp = xRoot;
                xRoot = yRoot;
                yRoot = tmp;
            }

            // まとめる先のグループの要素数を更新
            Parents[xRoot] += Parents[yRoot];

            // まとめられるグループのルートの親を書き換え
            Parents[yRoot] = xRoot;

            return true;
        }
        public bool IsUnited(long x, long y)
        {
            return Find(x) == Find(y);
        }
        public long Find(long x)
        {
            // 親がマイナスの場合は自分自身がルート
            if (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
            Parents[x] = Find(Parents[x]);

            return Parents[x];
        }
        public long GetSize(long x)
        {
            // ルート要素を取得して、サイズを取得して返す
            return -Parents[Find(x)];
        }
    }
}