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 JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

function main(input) {
	var lines = input.split('\n');
	var lines0 = lines[0].split(' ');
	var N = Number(lines0[0]);
	var M = Number(lines0[1]);
	var P = lines[1].split(' ');
	for(i=0;i<P.length;i++){
		P[i] = Number(P[i]);
	}
	var a = [];
	for(i=0;i<M;i++){
		a[i] = lines[i+2].split(' ');
		a[i][0] = Number(a[i][0]);
		a[i][1] = Number(a[i][1]);
	}
	var count = 0;
	for(i=0;i<N;i++){
		var f = true;
		var lastc = [];
		var c = [i+1];
		var c2 = [];
		var c3 = [i+1];
		var g = P[i];
		do{
			for(k=0;k<c.length;k++){
				for(j=0;j<M;j++){
					if(a[j][0] === c[k]){
						if(c.indexOf(a[j][1]) === -1 && c3.indexOf(a[j][1]) === -1){
							c2.push(a[j][1]);
							c3.push(a[j][1]);
						}
					} else if(a[j][1] === c[k]){
						if(c.indexOf(a[j][0]) === -1 && c3.indexOf(a[j][0]) === -1){
							c2.push(a[j][0]);
							c3.push(a[j][0]);
						}	
					}
					if(c.indexOf(g) >= 0){
						break;
					}
				}
				if(c.indexOf(g) >= 0){
					break;
				}
			}
			if(c.indexOf(g) >= 0){
				count += 1;
				f = false;
			}
			if(lastc === c){
				f = false;
			} else {
				c = c2;
				lastc = c;
			}
		}while(f===true);
	}
	console.log(count);
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));