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: Wrong Answer)

#include <iostream>
#include <algorithm>
using namespace std;

int par[100001];
int depth[100001];

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

int find(int x){
    if (par[x] == x){
        return x;
    }
    else{
        return par[x] = find(par[x]);
    }
}

void unite(int x, int y){
    x = find(x);
    y = find(y);
    if (x==y) return ;

    if(depth[x] < depth[y]){
        par[x] = y;
    }
    else{
        par[y] = x;
        if (depth[x] == depth[y]){
            depth[x]++;
        }
    }
}

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


int main(void){
    int N , M;
    int p[100002];

    cin >> N >> M;

    init(N);

    for(int i = 0; i < N ; i++){
        cin >> p[i+1];
    }

    int x, y;
    for(int i = 0; i < M ; i++){
        cin >> x >> y;
        unite(p[x+1], p[y+1]);
    }

    int ans=0;

    for(int i = 1; i < N+1 ; i++){
        if (same(p[i], i)){
            ans++;
        }
    }

    cout << ans << endl;
    return 0;
}