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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (N, M) = readLine()!!.split(" ").map { it.toInt() }
    val p = readLine()!!.split(" ").map { it.toInt() - 1 }.toIntArray()
    val unionFind = UnionFind(N)
    (1..M).forEach {
        val (x, y) = readLine()!!.split(" ").map { it.toInt() }
        unionFind.merge(x - 1, y - 1)
    }


    var answer = 0
    (0 until N).forEach {i ->
        // i and pi belongs same group, we can place i to the proper position
        if (unionFind.isSame(i, p[i])) {
            answer++
        }
    }
    println(answer)
}
class UnionFind(size: Int) {
    var parent: IntArray
    var height: IntArray
    var size: IntArray

    init {
        this.parent = IntArray(size)
        this.height = IntArray(size)
        this.size = IntArray(size)

        for (i in 0 until size) {
            makeSet(i)
        }
    }

    fun merge(x: Int, y: Int) {
        val rootX = root(x)
        val rootY = root(y)

        if (rootX == rootY) {
            return
        }

        if (height[rootX] > height[rootY]) {
            parent[rootY] = rootX
            size[rootX] += size[rootY]

        } else if (height[rootX] < height[rootY]) {
            parent[rootX] = rootY
            size[rootY] += size[rootX]

        } else {
            parent[rootY] = rootX
            size[rootX] += size[rootY]
            height[rootX]++
        }
    }

    fun isSame(x: Int, y: Int) : Boolean {
        return root(x) == root(y)
    }

    fun size(x: Int): Int {
        val rootX = root(x)
        return size[rootX]
    }

    private fun root(x: Int): Int {
        if (x != parent[x]) {
            parent[x] = root(parent[x])
        }
        return parent[x]
    }


    private fun makeSet(i: Int) {
        parent[i] = i
        height[i] = 0
        size[i] = 1
    }
}