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 PyPy3 (2.4.0) (judge verdict: Accepted)

#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = I()
    return l
def LIR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = LI()
    return l
def SR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = S()
    return l
def LSR(n):
    l = [None for i in range(n)]
    for i in range(n):l[i] = SR()
    return l
mod = 1000000007

#A
def A():
    return

#B
def B():
    return

#C
def C():
    s = input()
    k = I()
    l = []
    n = len(s)
    alp = list("abcdefghijklmnopqrstuvwxyz")
    for i in alp:
        for j in range(n):
            if s[j] == i:
                l.append(s[j])
                for m in range(j+1,min(j+k,n)):
                    l.append(s[j:m+1])
        l = sorted(list(set(l)))
        if len(l) >= k:break
    print(l[k-1])
#D
def D():
    def root(x):
        if par[x] == x:
            return x
        par[x] = root(par[x])
        return par[x]

    def same(x,y):
        return root(x) == root(y)

    def unite(x,y):
        x = root(x)
        y = root(y)
        if rank[x] < rank[y]:
            par[x] = y
        else:
            par[y] = x
            if rank[x] == rank[y]:
                rank[x] += 1

    n,m = LI()
    p = LI()
    for i in range(n):
        p[i] -= 1
    par = [i for i in range(n)]
    rank = [0 for i in range(n)]
    for i in range(m):
        x,y = LI()
        if not same(x-1,y-1):
            unite(x-1,y-1)
    for i in range(n):
        root(i)
    g = [[] for i in range(max(par)+1)]
    for i in range(n):
        g[par[i]].append(i)
        g[par[i]].append(p[i])
    ans = 0
    for i in g:
        i.sort()
        d = defaultdict(int)
        for j in i:
            d[j] += 1
            if d[j] == 2:
                ans += 1
    print(ans)
#E
def E():
    return

#F
def F():
    return

#G
def G():
    return

#H
def H():
    return

#Solve
if __name__ == "__main__":
    D()