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: Wrong Answer)

#include <algorithm>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <vector>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define show(x) cout << #x << " = " << x << endl;
using namespace std;
using ll = long long;
using pii = pair<int,int>;
#define MAX_V 100005
int V;
vector<int> G[MAX_V];
vector<int> rG[MAX_V];//辺を逆にしたグラフ
vector<int> vs;
bool used[MAX_V];
int cmp[MAX_V];//属する強連結成分のトポロジカル順序

void add_edge(int from, int to){
    G[from].push_back(to);
    rG[to].push_back(from);
}

void dfs(int v){
    used[v] = true;
    for(int i = 0; i < G[v].size(); ++i){
        if(!used[G[v][i]]) dfs(G[v][i]);
    }
    vs.push_back(v);
}

void rdfs(int v, int k){
    used[v] = true;
    cmp[v] = k;
    for(int i = 0; i < rG[v].size(); ++i){
        if(!used[rG[v][i]]) rdfs(rG[v][i], k);
    }
}
int scc(){
    memset(used, 0, sizeof(used));
    vs.clear();
    for(int v = 0; v < V; ++v){
        if(!used[v]) dfs(v);
    }
    memset(used, 0, sizeof(used));
    int k = 0;
    for(int i =(int) vs.size()-1; i >= 0; i--){
        if(!used[vs[i]]) rdfs(vs[i],k++);
    }
    return k;
}
int a[100005];
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m;
    cin >> n >> m;
    V = n;
    vector<int>p(n);
    rep(i,n){
        cin >> p[i];
        a[p[i]] = i;
    }
    vector<int>x(m),y(m);
    rep(i,m){
        cin >> x[i] >> y[i];
        x[i]--,y[i]--;
        add_edge(x[i],y[i]);
        add_edge(y[i],x[i]);
    }
    int k = scc();
   
    int ans = 0;
    rep(i,n){
        if(p[i] == i+1){
            ans++;
        }else{
            if(cmp[i] == cmp[a[p[i]-1]]){
                ans++;
            }
        }
    }
    cout << ans << endl;
    
}