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>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int n,m,p[100020],x[100020],y[100020],par[100020];

int find(int x){
    if (par[x]==x){
        return x;
    } else {
        return par[x]=find(par[x]);
    }
}

void unite(int x, int y){
    x=find(x);
    y=find(y);
    if(x==y){
        return ;
    }
    par[x]=y;
}

bool same(int x, int y){
    if (find(x)==find(y)){
        return true;
    } else {
        return false;
    }
}

int main(void){
    scanf("%d%d",&n,&m);
    for (int i=0; i<n; i++){
        scanf("%d",&p[i]);
        par[i]=i;
    }
    for (int i=0; i<m; i++){
        scanf("%d%d",&x[i],&y[i]);
        unite(x[i]-1,y[i]-1);
    }
    int ans=0;
    for (int i=0; i<n; i++){
        if (same(i,p[i]-1)==true){
            ans++;
        }
    }
    printf("%d\n",ans);
	return 0 ;
}