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: Wrong Answer)

#include <stdio.h>
#define limit 100000

long N,M,ans = 0;
long par[limit],rank[limit];

void init(long n){
  for(int i = 0;i < n;i++){
    par[i] = i;
    rank[i] = 0;
  }
}

long root(long x){
  if(par[x] == x) return x;
  else return par[x] = root(par[x]);
}

int same(long x,long y){
  if(root(x) == root(y)) return 1;
  else return 0;
}

void unite(long x,long y){
  x = root(x);
  y = root(y);
  if(x == y) return;

  if(rank[x] < rank[y]){
    par[x] = y;
  }else{
    par[y] = x;
    if(rank[x] == rank[y]) rank[x]++;
  }
}

int main(void){
  long x,y;
  scanf("%ld%ld",&N,&M);
  long p[N];

  init(N);
  for(int i = 0;i < N;i++){
    scanf("%ld",&p[i]);
  }
  for(int i = 0;i < M;i++){
    scanf("%ld%ld",&x,&y);
    x--; y--;
    unite(x,y);
  }

  for(int i = 0;i < N;i++){
    if(same(p[i],i+1)) ans++;
  }

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

  return 0;
}