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> P;
ll INFL = 1000000000000000010;//10^18 = 2^60
int INF = 2147483600;//10^9
ll MOD  = 1000000007;
int par[100010];
int Rank[100010];

void init(int n){
    for(int i = 0; i < n; i++){
        par[i] = i;
        Rank[i] = 1;
    }
}

int find(int x){ //根を返す関数
    if(par[x] == x) return x;
    else{
        return par[x] = find(par[x]);
    }
}

void unite(int x, int y){
    int rx = find(x), ry = find(y);
    if(Rank[rx] >= Rank[ry]){
        par[ry] = rx;
        if(Rank[rx] == Rank[ry]){
            Rank[rx]++;
        }
    }
    else{
        par[rx] = ry;
    }
    
}

bool same(int x, int y){
    return find(x) == find(y);
}




int main(void){
    int N,M;
    cin >> N >> M;
    vector<int> p(N);
    for(int i = 0; i < N; i++){
        cin >> p.at(i);
        p.at(i)--;
    }
    
    vector<int> x(M), y(M);
    for(int i = 0; i < M; i++){
        cin >> x.at(i) >> y.at(i);
        x.at(i)--;
        y.at(i)--;
    }
    
    init(N);
    for(int i = 0; i < M; i++){
        unite(x.at(i), y.at(i));
    }
    int ans = 0;
    for(int i = 0; i < N; i++){
        if(p.at(i) == i){
            ans++;
        }
        else if( same(p.at(i),  i) ){
            ans++;
        }
    }
    cout << ans << endl;
    
}