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 <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
#define mp  std::make_pair
#define ll long long int
#define abs(n) ((n>0)?(n):((n)*(-1)))
#define CEIL(a,b) (((a) + ((b)-1)) / (b))
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i,n) FOR(i,0,(n)-1)
#define VREP(v) for(std::vector<ll>::iterator itr = (v).begin(); itr != (v).end();itr++)
#define VALL(v) ((v).begin(),(v).end())
#define errln(v) cerr << (v) << endl;
#define println(v) cout << (v) << endl;

int p[100001];
int par[100001];

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

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

void unite(int x, int y){
	x = root(x);
	y = root(y);
	if(x == y) return;
	par[x] = y;
}

int main(int argc, const char * argv[]) {
	int N ,M ;
	cin >> N >> M;

	FOR(i,1,N){
		cin >> p[i];
		par[i] = i;
	}

	FOR(i,1,M){
		int x,y;
		cin >> x >> y;
		unite(x,y);
	}
	int ret = 0;
	FOR(i,1,N){
		if(same(i,p[i])) ret++;
	}
	println(ret);

	return 0;
}