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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <utility>
#include <string>
#include <stack>
#include <regex>
#include <numeric>
#include <queue>
#include <set>

#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;

struct union_find
{
private:
  vector<int> rank;
  void swap(int *x, int *y)
  {
    int tmp = *x;
    *x = *y;
    *y = tmp;
  }

public:
  vector<int> parent;
  int tree_num;
  void init(int nodenum)
  {
    parent.resize(nodenum, 0);
    for (int i = 0; i < nodenum; i++)
      parent[i] = i;
    rank.resize(nodenum, 0);
    tree_num = nodenum;
  }
  int find(int x)
  {
    if (parent[x] == x)
      return x;
    return parent[x] = find(parent[x]);
  }
  void unite(int x, int y)
  {
    x = find(x);
    y = find(y);
    if (x == y)
      return;
    if (rank[x] < rank[y])
      swap(&x, &y);
    rank[x]++;
    rank[y] = rank[x];
    parent[y] = x;
    tree_num--;
    return;
  }
};

int main(int argc, char const *argv[])
{
  ll n,m;
  cin >> n >> m;
  vector<ll> p(n);
  REP(i,n) {
    cin >> p[i];
    p[i]--;
  }
  union_find uf;
  uf.init(n);
  vector<pair<ll,ll>> xy(m+n);
  REP(i,m){
    ll x,y;
    cin >> x >> y;
    uf.unite(x-1,y-1);
    xy[i] = {x-1,y-1};
  }
  REP(i,n){
    xy[i+m] = {i,i};
  }
  std::map<ll,std::set<ll>> smap1;
  std::map<ll,std::set<ll>> smap2;
  for (auto ope : xy){
    ll x = ope.first;
    ll y = ope.second;
    ll root = uf.find(x);
    if (smap1.find(root) == smap1.end()) {
      smap1[root] = set<ll>();
      smap2[root] = set<ll>();
     }
      smap1[root].insert(x);
      smap1[root].insert(y);
      smap2[root].insert(p[x]);
      smap2[root].insert(p[y]);
  }
  ll ans = 0;
  for (auto itr = smap1.begin(); itr != smap1.end(); ++itr) {
    std::set<ll> res;
    auto mp2 = smap2[itr->first];
    std::set_intersection(ALL(itr->second),ALL(mp2),
    std::inserter(res,res.end()));
    ans += res.size();
  }
  cout << ans << endl;

  return 0;
}