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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>

using namespace std;

int F[200010], P[200010];
bool vis[200010];
int M, N;

int FindFather(int x) {
    return F[x] == x ? x : F[x] = FindFather(F[x]);
}

int Merge(int a, int b) {
    int Fa = FindFather(a),
        Fb = FindFather(b);

    if (Fa != Fb) F[Fb] = F[Fa];

    return 0;
}

int main()
{
    scanf("%d%d", &N, &M);

    for (int i = 1; i <= N; i++) {
        F[i] = i;
        vis[i] = false;
        scanf("%d", &P[i]);
    }

    for (int i = 0, u, v; i < M; i++) {
        scanf("%d%d", &u, &v);
        Merge(u, v);
    }

    int ans = 0;
    for (int i = 1; i <= N; i++) if (FindFather(i) == FindFather(P[i])) ans++;

    printf("%d\n", ans);

    return 0;
}