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<bits/stdc++.h>
using namespace std;
int bin[100005];
int du[100005];
int numid[100005];
int ic[100005];
int fin_bi(int id);
int main(void)
{
    for(int i = 0;i < 100005;i++)
        bin[i] = i,du[i] = 1;
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i = 1;i <= n;i++)
        scanf("%d",&numid[i]);
    for(int i = 0;i < m;i++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        int idx = fin_bi(x);
        int idy = fin_bi(y);
        if(idx != idy)
        {
            if(du[idx] > du[idy])
                du[idx] += du[idy],bin[idy] = idx;
            else 
            {
                du[idy] += du[idx];
                bin[idx] = idy;
            }
        }
    }
    int cnt = 0;
    for(int i = 1;i <= n;i++)
    {
       int icx = fin_bi(i);
       int icy = fin_bi(numid[i]);
       if(icx == icy)
       {
          cnt++;
       }
    }
    printf("%d\n",cnt);
    return 0;
}
int fin_bi(int id)
{
    while(bin[id] != id)
        id = bin[id];
    return id;
}