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>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
#include<limits.h>
#include<deque>
#include<cstdio>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1000000;
const int mod = 1000000007;
const int INF = 1 << 30;
#define rep(i,n) for(ll i=(ll)0;i<(ll)n;++i)
#define ALL(x) x.begin(),x.end()
#define pp pair<ll,ll>
#define fi first
#define se second
#define pb push_back
#define fix(n) fixed<<setprecision(n)
#define itr vector<ll>::iterator
int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
string YN(bool b) { return(b ? "YES" : "NO"); }
string yn(bool b) { return(b ? "Yes" : "No"); }
ll n, m, v[100010], a, b,gr[100010];
pp ps[100010];
vector<ll>edge[100010];
bool bfs(ll f, ll pos) {
	ll chec[100010] = {};
	queue<ll> q;
	q.push(pos);
	while (!q.empty()) {
		ll p = q.front(); q.pop();
		if (pos != p)edge[pos].pb(p);
		if (f == p)return 1;
		rep(i, edge[p].size())if (chec[edge[p][i]] == 0) { q.push(edge[p][i]); chec[edge[p][i]] = 1; }
	}
	return 0;
}
int main() {
	bool b = 0;
	ll ct = 0;
	cin >> n >> m;
	rep(i, n)gr[i] = INF;
	rep(i, n)cin >> v[i];
	rep(i, m) { cin >> ps[i].fi >> ps[i].se; --ps[i].first; --ps[i].second; if (ps[i].se < ps[i].fi)swap(ps[i].first, ps[i].se); }
	sort(ps, ps + m);
	while (!b) {
		b = 1;
		rep(i, m) {
			if (gr[ps[i].fi] == INF)gr[ps[i].fi] = i,b=0;
			ll abe = min(gr[ps[i].fi], gr[ps[i].se]);
			if (gr[ps[i].fi] != abe || gr[ps[i].se] != abe)b = 0;
			gr[ps[i].fi] = gr[ps[i].se] = min(gr[ps[i].fi], gr[ps[i].se]);
		}
	}
	rep(i, n) {
		if (gr[i] == gr[v[i] - 1])++ct;
	}
	cout << ct << endl;
	return 0;
}