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 <stdio.h>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set>
#include <iterator>
#pragma warning(disable:4996)

typedef long long ll;
#define MIN(a, b) ((a)>(b)? (b): (a))
#define MAX(a, b) ((a)<(b)? (b): (a))
#define MOD 1000000007
using namespace std;

long p[100005];
long flag[100005];
vector<vector<long> > a;

void Get( long i, vector<long>& vv )
{
    if(flag[i]) return;
    
    flag[i]=1;
    vv.push_back(i);
    long k;
    for(k=0; k<(long)a[i].size(); k++) {
        Get(a[i][k], vv);
    }

    return;
}

int main(int argc, char* argv[])
{
    long n, m;
    scanf("%ld %ld", &n, &m);
    long i;
    for(i=0; i<n; i++) {
        scanf("%ld", &(p[i]));
        p[i]--;
    }

    a.resize(n);
    for(i=0; i<m; i++) {
        long x, y;
        scanf("%ld %ld", &x, &y);
        a[x-1].push_back(y-1);
        a[y-1].push_back(x-1);
    }
    
    long ans=0;
    for(i=0; i<n; i++) {
        if(flag[i]) continue;
        vector<long> tmp;
        Get(i, tmp);
        
        sort(tmp.begin(), tmp.end());
        vector<long> tmp2(tmp.size());
        long k;
        for(k=0; k<(long)tmp.size(); k++) {
            tmp2[k]=p[tmp[k]];
        }
        sort(tmp2.begin(), tmp2.end());

        vector<long> tmp3;
        set_intersection(tmp.begin(), tmp.end(), tmp2.begin(), tmp2.end(), back_inserter(tmp3));
        ans += (long)tmp3.size();
    }

    printf("%ld\n", ans);

    return 0;
}