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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include<string>
#include<cmath>

using namespace std;

#define ll long long int
#define ld long double
#define loop(i, n) for (int i = 0; i < n; i++)
#define loops(i, s, n) for (int i = s; i < n; i++)
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define sz(x) ((int)(x).size())
#define fi first
#define se second
#define deb(x) cout<< #x << '=' << x <<endl
#define MOD 1000000007
const int N = 1e5 + 5;
int n,m;
vector<int> adj[N];
int a[N];
int vis[N];
bool dfs(int u , int target){
    vis[u] = 1;
    if(u == target){
        return true;
    }
    for(auto v : adj[u]){
        if(!vis[v]){
            if(dfs(v , target)){
                return true;
            }
        }
    }
    return false;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin>>n>>m;
    for(int i = 1; i <= n; i++){
        cin>>a[i];
    }
    for(int i = 0; i < m; i++){
        int x , y;
        cin>>x>>y;
        adj[x].pb(y);
        adj[y].pb(x);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++){
        if(a[i] != i){
        memset(vis , 0 , sizeof(vis));
        bool ok = dfs(a[i] , i);
            if(ok){
                ans++;
            }
        }else{
            ans++;
        }
    }
    cout<<ans;
}