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 rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()

using namespace std;
typedef long long ll; typedef vector<int> VI; typedef pair<int,int> pii;
const int MD = (int)1e9 + 7; typedef vector<ll> VL;
void dbg(){ cerr << "\n"; }  template <class T, class ...T2> void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }
template <class T, class T2> void amax(T& a,T2 b){ if(a < b) a = b;}

struct UF{
	vector<int> p;
	UF() {}
	UF(int n) : p(n, -1) {}
	int find(int a){ return p[a] < 0 ? a : p[a] = find(p[a]); }
	bool unite(int a, int b) {
		a = find(a); b = find(b);
		if(a == b) return false;
		if(-p[a] < -p[b]) swap(a, b);
		p[a] += p[b]; p[b] = a;
		return true;
	}
	bool same(int a, int b){ return find(a) == find(b); }
};

int main()
{
	cin.tie(0); ios_base::sync_with_stdio(false);
	int n, m;
	cin >>n >>m;
	VI p(n);
	rep(i, n){ cin >>p[i]; p[i]--;}
	UF ufa(n);
	rep(i, m){
		int x, y;
		cin >>x >>y;
		ufa.unite(x-1, y-1);
	}
	int ans = 0;
	rep(i, n){
		if(ufa.same(i, p[i])) ans++;
	}
	cout <<ans <<"\n";
	return 0;
}