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 namespace std;

template <typename T> T gcd(T x, T y) {
  if (y == 0)
    return x;
  return gcd(y, x % y);
}

template <typename T> T lcm(T x, T y) {
  if (x == 0 || y == 0)
    return 0;
  return x / gcd(x, y) * y;
}

vector<int> p;
vector<int> r;

void init(int n) {
  p.reserve(n);
  r.resize(n);
  for (int i = 0; i < n; i++) {
    p[i] = i;
    r[i] = 0;
  }
}

int root(int x) { return p[x] == x ? x : p[x] = root(p[x]); }

bool same(int x, int y) { return root(x) == root(y); }

void unite(int x, int y) {
  x = root(x);
  y = root(y);
  if (x == y)
    return;

  if (r[x] < r[y])
    p[x] = y;
  else {
    p[y] = x;
    if (r[x] == r[y])
      r[x]++;
  }
}

int main() {
  int N, M;
  cin >> N >> M;
  init(N + 1);
  vector<int> p(N);
  for (auto &x : p)
    cin >> x;

  for (int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    unite(x, y);
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    if (same(i + 1, p[i]))
      ans++;
  }
  cout << ans << endl;
}