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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

class Main{


	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] p = new int[n];
		for(int i = 0;i < n;i++){
			p[i] = sc.nextInt();
		}
		UnionFind tree = new UnionFind(n+1);
		for(int i = 0;i < m;i++){
			int x = sc.nextInt();
			int y = sc.nextInt();
			tree.unite(x, y);
		}
		int ans = 0;
		for(int i = 0;i < n;i++){
			if(p[i] == i+1 || tree.same(p[i], i+1)){
				ans++;
			}
		}
		System.out.println(ans);
		
		






	}
}

class UnionFind{
	int[] parent;
	public UnionFind(int N){
		this.parent = new int[N];
		for(int i = 0; i < N;i++){
			parent[i] = i;
		}
	}
	public int root(int x){			//根要素を取得
		if(parent[x] == x){			//初期状態
			return x;
		}else{
			return parent[x] = root(parent[x]);    	//再帰によって根を取得
		}
	}
	public void unite(int x,int y){ 	//統合メソッド
		if(same(x,y)){					//同じ根なら終了
			return;
		}
		int xroot = root(x);
		int yroot = root(y);
		parent[xroot] = yroot;			//統合
		return;

	}

	public boolean same(int x,int y){			//根が同じか（同じグループに属しているか）を判定
		return(root(x) == root(y));
	}
}



class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}