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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int par[100001];//親
int siz[100000];//サイズ

void ini(int n){ //初期化
  for(int i=0;i<=n;i++){
    par[i] = i;
    siz[i] = 1;
  }
}
int root(int x){ //根を返す
  if(par[x] == x) return x;
  else return par[x] = root(par[x]);
}

void uni(int x, int y){ //くっつける
  x = root(x);
  y = root(y);
  if(x != y){
    siz[x] += siz[y];
    par[y] = x;
  }
}

int same(int x,int y){
  return root(x) == root(y);
}

int size(int x){ //サイズを返す
  return siz[root(x)];
}

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