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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using ll = long long;
using namespace std;
const int INFint = 2e9+1;
const ll INFll = 2e18+1;
ll MOD=1e9+7;

int main(){
  int N,M;
  cin>>N>>M;
  vector<int> p(N);
  for(int i(0);i<N;i++){
    cin>>p[i];
  }
  vector<int> group(N+1);
  vector<int> G[N+1];
  for(int i(0);i<M;i++){
    int x,y;
    cin>>x>>y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  bool pathed[100001] = {0};
  int cnt(1);
  for(int i(1);i<=N;i++){
    if(pathed[i]) continue;
    cnt++;
    queue<int> q;
    q.push(i);
    while (!q.empty()){
      int now = q.front();q.pop();
      pathed[now] = 1;
      group[now] = i;
      for(auto next:G[now]){
        if(!pathed[next]){
          q.push(next);
        }
      }
    }
  }
  for(int i(1);i<=N;i++){
    //cout << group[i] <<" ";
  }
  int ans(0);
  for(int i(1);i<=N;i++){
    //cout << group[p[i]] << " " << group[i] << endl;
    if (group[p[i-1]] == group[i]){
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}