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: Runtime Error)

#include "iostream"
#include "vector"
#include "string"
#include "algorithm"
#include "math.h"
#include "cstring"
#include "stack"
#include "queue"

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<vector<int> > mat;

//UnionFind
const int MAX_N = 10100;
const int MAX_Q = 10100;
struct UnionFind
{
  int par[MAX_N];	//親
  int height[MAX_N];	//木の深さ
  int size[MAX_N];	//子の数

	//n要素で初期化
  void init (int n)
  {
    for (int i = 0; i < n; i++)
    {
      par[i] = i;
      height[i] = 0;
      size[i] = 1;
    }
  }

	//木の根を求める
  int find (int x)
  {
    if (par[x] == x)
    {
      return x;
    }
    else
    {
      return par[x] = find(par[x]);
    }
  }

	//xとyの属する集合を併合
  void unite (int x, int y)
  {
    x = find(x);
    y = find(y);
    if (x == y)
    {
      return;
    }

    if (height[x] < height[y])
    {
      par[x] = y;
      size[y] = size[x] + size[y];
    }
    else
    {
      par[y] = x;
      size[x] = size[x] + size[y];
      if (height[x] == height[y])
      {
        height[x]++;
      }
    }
  }

	//xとyが同じ集合に属するか否か
  bool same (int x, int y)
  {
    return find(x) == find(y);
  }
};

int main()
{
  int n, m;
  cin >> n >> m;
  int p[MAX_N];
  for (int i = 0; i < n; i++)
  {
    cin >> p[i];
    p[i]--;
  }
  UnionFind uf;
  uf.init(n);
  int x[MAX_Q], y[MAX_Q];
  for (int i = 0; i < m; i++)
  {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    uf.unite(x[i], y[i]);
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
  {
    if (uf.same(i, p[i]))
    {
      ans++;
    }
  }
  cout << ans << "\n";
}