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>
#define MAXN 100000

std::vector <int> G[1 + MAXN];
int p[1 + MAXN];
int seen[1 + MAXN];
std::vector <int> V;
std::map <int, int> M;

void dfs(int node){
    V.push_back(node);
    M[p[node]]++;
    for(auto y: G[node]) if(!seen[y]){
        seen[y] = 1;
        dfs(y);
    }
}

int main(){
    FILE*fi,*fo;
    //fi = fopen("a.in","r");
    //fo = fopen("a.out","w");
    fi = stdin;
    fo = stdout;

    int n, m;
    fscanf(fi,"%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        fscanf(fi,"%d", &p[i]);

    for(int i = 1; i <= m; i++){
        int x, y;
        fscanf(fi,"%d%d", &x, &y);
        G[x].push_back(y);
        G[y].push_back(x);
    }

    int con = 0;
    for(int i = 1; i <= n; i++)
        if(!seen[i]){
            seen[i] = 1;
            dfs(i);
            for(auto y: V)
                if(M[y]) con++;
            for(auto y: V)
                M[p[y]]--;
            V.clear();
        }
    fprintf(fo,"%d", con);

    fclose(fi);
    fclose(fo);
    return 0;
}