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 <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

const int nmax = 100005;
vector<int> g[nmax];
int v[nmax];
bool visited[nmax];
vector<int> pos;
vector<int> val;

void dfs(int node) {
    pos.push_back(node);
    val.push_back(v[node]);
    for(int it : g[node]) {
        if(!visited[it]) {
            visited[it] = 1;
            dfs(it);
        }
    }
}

int main() {
    int n,m;
    cin >> n >> m;
    for(int i = 1;i <= n;i ++)
        cin >> v[i];
    for(int i = 1;i <= m;i ++) {
        int x,y;
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    int nsol = 0;
    for(int i = 1;i <= n;i ++) {
        if(visited[i] == 0) {
            pos.clear();
            val.clear();
            visited[i] = 1;
            dfs(i);
            sort(pos.begin(),pos.end());
            sort(val.begin(),val.end());
            int ii = 0,j = 0;

            while(ii < pos.size() && j < val.size()) {
                if(pos[ii] == val[j]) {
                    ii ++;
                    j ++;
                    nsol ++;
                } else if(pos[ii] > val[j])
                    j ++;
                else if(pos[ii] < val[j])
                    ii ++;
            }
        }
    }
    cout << nsol;
    return 0;
}