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)

#include <stdio.h>
#include <stdlib.h>

typedef struct element {
    struct element *next;
} element;

typedef struct {
    int n;
    element *head;
    element *tail;
} set;

int *l;
set **f;

element *elements;

void initialize(int n) {
    int i;

    l = calloc(n + 1, sizeof(int));
    f = calloc(n + 1, sizeof(set *));
    elements = calloc(n + 1, sizeof(element));

    for (i = 1; i <= n; i++) {
        f[i] = NULL;
    }
}

void new_pair(int x, int y) {
    set *s = malloc(sizeof(set));

    f[x] = f[y] = s;

    elements[x].next = &elements[y];
    elements[y].next = NULL;

    s->n = 2;
    s->head = &elements[x];
    s->tail = &elements[y];
}

void merge(set *s1, set *s2) {
    element *e;

    if (s1->n < s2->n) {
        merge(s2, s1);
        return;
    }

    /** s2が小さいのでs1に寄せる */
    for (e = s2->head; e; e = e->next) {
        f[e - elements] = s1;
    }

    s1->n += s2->n;
    s1->tail->next = s2->head;
    s1->tail = s2->tail;
}

void insert(int x, set *s) {
    f[x] = s;
    element *e = &elements[x];
    e->next = s->head;
    s->head = e;
    s->n++;
}

void add_pair(int x, int y) {
    set *f_x, *f_y;

    if ((f_x = f[x])) {
        if ((f_y = f[y])) {
            if (f_x != f_y) {
                merge(f_x, f_y);
            }
        } else {
            insert(y, f_x);
        }
    } else {
        if ((f_y = f[y])) {
            insert(x, f_y);
        } else {
            new_pair(x, y);
        }
    }
}

int movable(int from, int to) {
    if (from == to) {
        return !0;
    }

    return f[from] && f[to] && f[from] == f[to];
}

int main()
{
    int i;
    int n, m;
    int x, y;
    int r;

    scanf("%d %d", &n, &m);

    initialize(n);

    for (i = 1; i <= n; i++) {
        scanf("%d", &l[i]);
    }

    for (i = 1; i <= m; i++) {
        scanf("%d %d", &x, &y);
        add_pair(x, y);
    }

    r = 0;
    for (i = 1; i <= n; i++) {
        if (movable(i, l[i])) {
            r++;
        }
    }

    printf("%d\n", r);
    return 0;
}