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 (Clang 3.8.0) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<set>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)
#define repn(i, n) for(int i = 1; i <= n; i++)
#define repr(i, n) for(int i = n-1; i >= 0; i--)
#define all(x) (x).begin(), (x).end()
#define lint long long

bool seen[100005];

int main(){
    int n, m; cin >> n >> m;
    int p[n], num[n];
    rep(i, n){
        int x; cin >> x; x--;
        p[i] = x;
        num[x] = i;
    }

    vector<int> to[100005];
    rep(i, m){
        int x, y; cin >> x >> y;
        x--; y--;
        to[x].push_back(y);
        to[y].push_back(x);
    }

    int res = 0;
    rep(i, n){
        if(seen[i]) continue;
        set<int> con;

        queue<int> q;
        q.push(i);
        while(!q.empty()){
            int x = q.front(); q.pop();
            seen[x] = true;
            con.insert(x);
            for(int& y : to[x]){
                if(!seen[y]) q.push(y);
            }
        }

        auto itr = con.begin();
        while(itr != con.end()){
            int x = *itr;
            if(con.find(num[x]) != con.end()) res++;
            itr++;
        }
    }
    cout << res << endl;
}