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>
//小さい順にソート
int c[200002][2];
void merge(int i,int j,int s[][2]){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
      else if(j<b){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else if(s[a][0]<s[b][0] || (s[a][0]==s[b][0] && s[a][1]<s[b][1])){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else{
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
    }
    for(d=i;d<=j;d++){
      s[d][0]=c[d][0];
      s[d][1]=c[d][1];
    }
  }
}

int main(){
  int n,m,i;
  scanf("%d%d",&n,&m);
  int p[n+1],x[2*m+1][2],mass[n+1];
  for(i=1;i<=n;i++){
    scanf("%d",&p[i]);
    mass[i]=i;
  }
  for(i=1;i<=m;i++){
    scanf("%d%d",&x[i][0],&x[i][1]);
    x[i+m][0]=x[i][1];
    x[i+m][1]=x[i][0];
  }
  merge(1,2*m,x);
  for(i=1;i<=2*m;i++)
    if(mass[x[i][1]]>mass[x[i][0]])
      mass[x[i][1]]=mass[x[i][0]];
    else
      mass[x[i][0]]=mass[x[i][1]];
  int count=0;
  for(i=1;i<=n;i++)
    if(mass[i]==mass[p[i]])
      count++;
  printf("%d\n",count);
  return 0;
}