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;

#define rep(i, n) for(int i = 0; i < (int)n; ++i)
#define FOR(i, a, b) for(int i = a; i < (int)b; ++i)
#define rrep(i, n) for(int i = ((int)n - 1); i >= 0; --i)

typedef long long ll;
typedef long double ld;

const int Inf = 1e9;
const double EPS = 1e-9;
const int MOD = 1e9 + 7;

class UnionFind {
    public:
        vector<int> rank, p, size;

        UnionFind() {}
        UnionFind(int s) {
            rank.resize(s, 0);
            p.resize(s, 0);
            size.resize(s, 0);
            rep (i, s) init(i);
        }

        void init(int x) {
            p[x] = x;
            rank[x] = 0;
            size[x] = 1;
        }

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

        void unite(int x, int y) {
            if (isSame(x, y)) return;
            link(root(x), root(y));
        }

        void link(int x, int y) {
            if (rank[x] > rank[y]) {
                p[y] = x;
                size[x] += size[y];
            } else {
                p[x] = y;
                size[y] += size[x];
                if (rank[x] == rank[y]) {
                    rank[y]++;
                }
            }
        }

        int root(int x) {
            if (x != p[x]) {
                p[x] = root(p[x]);
            }
            return p[x];
        }

        int getSize(int x) {
            return size[root(x)];
        }
};

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    UnionFind uf = UnionFind(n);
    vector<int> p(n);
    rep (i, n) {
        cin >> p[i];
        p[i]--;
    }
    rep (i, m) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        uf.unite(x, y);
    }
    int res = 0;
    rep (i, n) if (uf.isSame(i, p[i])) res++;
    cout << res << endl;
    
    return 0;
}