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;

using ll = long long;
#define ALL(c) (c).begin(), (c).end()
const int MOD = 1000000007;

class UnionFind {
    vector<int> parent;

    public:
        UnionFind(int size) {
            parent = vector<int>(size, -1);
        }

        int find(int x) {
            if (parent[x] < 0) {
                // parent itself
                return x;
            } else {
                return parent[x] = find(parent[x]);
            }
        }

        bool unite(int x, int y) {
            int rX, rY;
            if ((rX = find(x)) == (rY = find(y))) {
                // already united
                return false;
            }
            if (size(rX) < size(rY)) {
                int tmp = rX;
                rX = rY;
                rY = tmp;
            }
            parent[rX] += parent[rY];
            parent[rY] = rX;
            return true;
        }

        int size(int x) {
            return -parent[find(x)];
        }

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

        int component_count() {
            int c = 0;
            for (const int x : parent) {
                if (x < 0) {
                    c++;
                }
            }
            return c;
        }
};

int main() {
    int n, m;
    cin >> n >> m;

    vector<int> p(n);
    for (int i = 0; i < n; i++) {
        cin >> p[i];
        p[i]--;
    }

    UnionFind uf = UnionFind(n);
    int x, y;
    for (int i = 0; i < m; i++) {
        cin >> x >> y;
        uf.unite(x - 1, y - 1);
    }

    int t = 0;
    for (int i = 0; i < n; i++) {
        if (uf.isSame(i, p[i])) {
            t++;
        }
    }

    cout << t << endl;
}