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>
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
 
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define RFOR(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)
#define VSORT(v) sort(v.begin(), v.end())
#define SORT(v, n) sort(v, v+n)
#define VREV(v,n) reverse(v,v+n)
#define REV(v,n) reverse(v,v+n)
#define vi vector<int>
#define pf push_front
#define pb push_back

template<class T, size_t N> size_t countof(const T (&array)[N]) { return N; }
template <class T> void chmin(T&a,const T&b) { a = min(a,b); }
template <class T> void chmax(T&a,const T&b) { a = max(a,b); }
 
void print(){cout<<"\n";}
template <class Head, class... Tail>
void print(Head&& h,Tail&&... t){ 
	if(sizeof...(t)==0)
		cout<<h;
	else
		cout<<h<<' ';
	print(move(t)...);
}

void mprint(auto *matrix,auto N){
	REP(i,N) cout<<matrix[i]<<(i!=N-1?' ':'\n');
	return;
}

const double EPS =1e-9;
const int INF =2e+9;
const ll INFL =9e+18;
const long MOD =1e+9+7;
#define PI 3.14159265258979

#define P pair<int,int>
#define Pll pair<ll,ll>
#define iP pair<int,P>
int dy[]={0, 0, 1, -1, 1, 1, -1, -1};
int dx[]={1, -1, 0, 0, 1, -1, -1, 1};

inline int in(){int x; cin>>x; return x;}

struct UnionFind{
	vector<int> par; //データiが属する親の番号
	vector<int> sizes; //根ノードiの木に含まれるデータの数

	// 最初は全てのデータiがグループiに存在するものとして初期化
	UnionFind(int n):par(n),sizes(n, 1){
		REP(i,n) par[i] = i;
	}

	// 根を張り替えながら再帰的に根ノードを探す
	int find(int x){
		if(x==par[x]) return x;
		return par[x]=find(par[x]);
	}

	// 2つのデータx, yが属する木をマージする
	void unite(int x, int y) {
		// データの根ノードを得る
		x = find(x);
		y = find(y);

		// 既に同じ木に属しているならマージしない
		if (x == y) return;

		// xの木がyの木より大きくなるようにする
		if (sizes[x] < sizes[y]) swap(x, y);

		// xがyの親になるように連結する
		par[y] = x;
		sizes[x] += sizes[y];
		sizes[y] = 0;  // sizes[y]は無意味な値となるため
	}

	// 2つのデータx, yが属する木が同じならtrueを返す
	bool same(int x, int y) {
		return find(x) == find(y);
	}

	// データxが含まれる木の大きさを返す
	int size(int x) {
		return sizes[find(x)];
	}
};

int main(void){
	cin.tie(0);
	ios_base::sync_with_stdio(false);

	int N,M,x,y,p[100000];
	UnionFind uf(100001);
	set<int> st[100001];

	cin>>N>>M;
	REP(i,N) cin>>p[i];
	REP(j,M){
		cin>>x>>y;
		uf.unite(x,y);
	}
	
	FOR(i,1,N+1) st[uf.find(i)].insert(p[i-1]);
	
	int result=0;
	FOR(i,1,N+1){
		int tem=uf.find(i);
		set<int>::iterator it = st[tem].find(i);
		if(it != st[tem].end()) result++;
	}
	print(result);
	return 0;
}