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: Runtime Error)

#include <stdio.h>
#define N 10001
#define M 10001

int main(void){
  int i,j,t,n,m,x,y,g;
  int cnt;
  int p[N];
  int uf[N+1];
  int link[N+1];
  
  scanf("%d %d",&n,&m);
  for( i=1; i<=n; i++){
    scanf("%d",&p[i]);
    uf[i] = 0;
  }
  for( g=1, i=0; i<m; i++){
    scanf("%d %d",&x,&y);
    /* Found New Union */
    if( uf[x]==0 && uf[y]==0 ){
      uf[x] = g;
      uf[y] = g;
      link[g] = g;
      g++;
    }
    else if( uf[x] == 0 ){
      uf[x] = uf[y];
      link[ uf[x] ] = link[ uf[y] ];
    }
    else if( uf[y] == 0 ) {
      uf[y] = uf[x];
      link[ uf[y] ] = link[ uf[x] ];
    }
    /* Founf Diffrent Union */
    else link[ uf[y] ] = link[ uf[x] ];
  }
  
  /* Numbering Independ Tree */
  for( i=1; i<=n; i++){
    if( uf[i] == 0 ){
      uf[i] = g;
      link[g] = g;
      g++;
    }
  }
  
  for( cnt=0, i=1; i<=n; i++){
    //printf("%d : %d\n", i,link[ uf[i] ]);
    if( link[ uf[i] ] == link[ uf[p[i-1]] ]) cnt++;
  }
  
  printf("%d\n",cnt);
  
  return 0;
}