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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define vi vector<int>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi 3.141592653589793238
#define rep(i,a,b) for (int i = a; i <= b; i++)
#define zip(i,a,b) for(int i=a;i<b;i++)
#define rzip(i,a,b) for(int i=a;i>=b;i--)
#define ll unsigned int
#define test int t;cin>>t; while(t--)
#define en '\n'
typedef pair<int, int>     pii;
typedef pair<char, int>    pci;
typedef pair<char, char>   pcc;
typedef vector<pii>       vii;
typedef long double ld;
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define sz(x) (int)x.size()
#define INF (1e18+5)
#define inf (1e9+5)
#define mod 1000000007


void __print(int x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
// You should only debug a pair of simple data types. For example,
// the debug won't work if one of pair's elements is collection type
// (std::vector, std::map, std::set...).
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif


// bool check_prime(int n)  return whether a number is prime or not O(sqrt(N));
// int BE(int x,int n,int m)  return x^n%m; O(logN);
// void sieve()  from number 1-1000001 saare prime store kar lega  in bool is_prime[1000001] array
// vector<int> z_array(string s)  return vector which is Z-array of string s;
// vector<int> lps(string s)      return vector which is lps array of string s;
// int power(int x,int n) return x^n; O(logN);
// using_ordered_set (template for using ordered set , replace first 2 lines of this page with this code..);
const int gf = 1e6 + 9;
vi adj[gf];
bool vis[gf];
map<int, int>m1;
int cnt;

void dfs(int u)
{
	vis[u] = 1;
	m1[u] = cnt;
	for (int x : adj[u])
		if (!vis[x])
			dfs(x);
}

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(0);
#ifndef ONLINE_JUDGE
	if (fopen("input.txt", "r"))
	{
		freopen ("input.txt" , "r" , stdin);
		freopen ("output.txt" , "w" , stdout);
	}
#endif
	//_______________________________-code starts-_______________________________________________

	int n, m;
	cin >> n >> m;
	int p[n + 1];
	rep(i, 1, n)
	{
		cin >> p[i];
	}

	//int m;
	//cin >> m;
	zip(i, 0, m)
	{
		int x, y;
		cin >> x >> y;
		adj[x].pb(y);
		adj[y].pb(x);
	}
	int ans = 0;
	cnt++;

	rep(i, 1, n)
	{
		if (!vis[i])
		{
			dfs(i);
			cnt++;
		}
	}
	rep(i, 1, n)
	{
		if (m1[i] == m1[p[i]])
			ans++;
	}
	cout << ans << en;



	return 0;
}