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 (GCC 5.4.1) (judge verdict: Accepted)

/* Coached by rainboy */
#include <stdio.h>
#include <string.h>

#define N 100000
#define M 100000

int oo[1 + M * 2], oj[1 + M * 2];

int link(int o, int j) {
	static int _ = 1;

	oo[_] = o, oj[_] = j;
	return _++;
}

int ae[N], cc[N];
char visited[N];

void dfs(int i, int c) {
	int o;

	if (visited[i])
		return;
	visited[i] = 1;
	cc[i] = c;
	for (o = ae[i]; o; o = oo[o])
		dfs(oj[o], c);
}

int main() {
	static int ii[N + 1];
	int n, m, i, p, c, ans;

	scanf("%d%d", &n, &m);
	for (i = 0; i < n; i++) {
		scanf("%d", &p);
		ii[p] = i;
	}
	while (m--) {
		int x, y;

		scanf("%d%d", &x, &y), x--, y--;
		ae[x] = link(ae[x], y);
		ae[y] = link(ae[y], x);
	}
	memset(cc, -1, n * sizeof *cc);
	c = 0;
	for (i = 0; i < n; i++)
		if (!visited[i])
			dfs(i, c++);
	ans = 0;
	for (p = 1; p <= n; p++)
		if (cc[ii[p]] == cc[p - 1])
			ans++;
	printf("%d\n", ans);
	return 0;
}