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++ (GCC 9.2.1) (judge verdict: Accepted)

    #include<iostream>
    #include<algorithm>
    #include<functional>
    #include<cmath>
    #include<string>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<map>
    #include<set>
    #include<deque>
    using namespace std;
    #define ll long long
    const int mod = 1000000007;
    const ll INF = 1000000000000000000;
     
    struct UnionFind {
    	vector<int> par;
    	UnionFind(int x) : par(x, -1) {}
    	void init(int x) { par.assign(x, -1); }
    	int root(int x) {
    		if (par[x] <= -1) return x;
    		else return root(par[x]);
    	}
    	bool issame(int x, int y) {
    		return root(x) == root(y);
    	}
    	bool merge(int x, int y) {
    		x = root(x); y = root(y);
    		if (x == y) return false;
    		if (par[x] > par[y]) swap(x, y);
    		par[x] += par[y];
    		par[y] = x;
    		return true;
    	}
    	int size(int x) {
    		return -par[root(x)];
    	}
    };
     
    int p[100010];
    int main()
    {
    	int N, M;
    	cin >> N >> M;
    	for (int i = 0; i < N; i++) {
    		cin >> p[i];
    		p[i]--;
    	}
    	UnionFind uf(N);
    	for (int i = 0; i < M; i++) {
    		int a, b;
    		cin >> a >> b;
    		a--; b--;
    		uf.merge(a, b);
    	}
    	int ans = 0;
    	for (int i = 0; i < N; i++) {
    		if (uf.issame(i, p[i])) ans++;
    	}
    	cout << ans << endl;
    }