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>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define rep1(i,n) for(int i=1; i<=(n); ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define sz(v) (int)v.size()
#define pb push_back
using namespace std;
struct UnionFind {
  vector<int> par;
  vector<int> rank;
  vector<int> num;
  
  UnionFind(int N) : par(N),rank(N),num(N) {
        for(int i = 0; i < N; i++){ 
          par[i] = i;
          rank[i] = 0;
          num[i] = 1;
        }
   }
      
  void init(int N){
        for(int i = 0; i < N; i++){ 
          par[i] = i;
          rank[i] = 0;
          num[i] = 1;
        }
  }
  
  int root(int x) {
    if (par[x] == x) return x;
    return par[x] = root(par[x]);
  }
  
  int ranker(int x){
    x = root(x);
    return rank[x];
  }
  
  int number(int x){
    x = root(x);
    return num[x];
   }
  
  void unite(int x, int y){
    int rx = root(x);
    int ry = root(y);
    if ( rx == ry ) return;
    if ( rank[rx] < rank[ry] ){
      par[rx] = ry;
      num[ry] += num[rx];
      num[rx] = 0;
    } else {
      par[ry] = rx;
      num[rx] += num[ry];
      num[ry] = 0;
      if ( rank[rx] == rank[ry] ) rank[rx]++;
    }
   }
  
  bool same(int x, int y){
    return root(x) == root(y);
  }
};

int N,M,temp,x,y;

int main(){
    cin >> N >> M;
    UnionFind tree(N+5);
    vector<int> p(N+1);
    rep1(i,N) cin >> p[i];
    rep(i,M){
        cin >> x >> y;
        tree.unite(x,y);
    }
    int ct = 0;
    rep1(i,N) if(tree.same(i,p.at(i))) ct++;
    cout << ct << endl;
    
}