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;
typedef long long ll;
typedef pair<int,int> pii;

class UnionFind{
public:
    UnionFind(int n){
        size = n;
        parent = vector<int>(size,-1);
    }
    int getSize(){return size;}
    int getParentElement(int index){return parent[index];}
    void printParent(){
        for(auto v: parent) {cout << v << " ";}
        cout << endl;
    }
    int find(int x){
        int a = x;
        vector<int> children;
        while(parent[a]>=0){
            children.push_back(a);
            a = parent[a];
        }
        for(auto child: children){parent[child] = a;}
        return a;
    }
    int Union(int x,int y){
        int root_x = find(x);
        int root_y = find(y);
        if(root_x != root_y){
            if(parent[root_x]>=parent[root_y]){
                parent[root_x] += parent[root_y];
                parent[root_y] = root_x;
            }else{
                parent[root_y] += parent[root_x];
                parent[root_x] = root_y;
            }
        }
        return parent[root_x];
    }
private:
    int size;
    vector<int> parent;
};

int main(){
    int n, m;
    cin >> n >> m;
    vector<int> p(n);
    vector<pii> xy(m);
    UnionFind uf(n);
    for(int i = 0; i < n; i++){
        int a;
        cin >> a;
        a--;
        p[i] = a;
    }
    for(int i = 0; i < m; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        uf.Union(a,b);
    }
    int ans = 0;
    for(int i = 0; i < n; i++){
        if(uf.find(i)==uf.find(p[i])){ans++;}
    }

    cout << ans << endl;



    return 0;

}