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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for (int i = x; i < n; i++)
#define OP(m) cout << m << endl
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int const N = 50001;

/*void dfs(int u, int c)
{
    connect[u] = c;
    rep(i, G[u].size())
    {
        if (connect[G[u][i]] == -1)
        {
            dfs(G[u][i], c);
        }
    }
    return;
}*/

int main()
{
    int n, m;
    cin >> n >> m;
    int p[n] = {};
    vector<int> G[n];
    vector<int> connect(n, -1);
    rep(i, n)
    {
        cin >> p[i];
        p[i]--;
    }

    rep(i, n)
        connect[i] = -1;

    rep(i, m)
    {
        int x, y;
        cin >> x >> y;
        x--, y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    int cnt = 0;

    rep(u, n)
    {
        if (connect[u] != -1)
            continue;
        queue<int> q;
        q.push(u);
        cnt++;
        while (!q.empty())
        {

            int v = q.front();
            connect[v] = cnt;
            q.pop();
            rep(i, G[v].size())
            {
                if (connect[G[v][i]] == -1)
                    q.push(G[v][i]);
            }
        }
    }

    int ans = 0;
    rep(i, n) if (connect[i] == connect[p[i]]) ans++;

    OP(ans);

    return 0;
}