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)

/* ABC065 D - 連結 / Connectivity */
#include <iostream>
#include <set>
#include <queue>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <climits>
#include <sstream>
#include <iomanip>
#include <map>
#include <stack>
#include <numeric>

using namespace std;

/*-----------------------------------------------------------------------------
　定義
 -------------------------------------------------------------------------------*/
#define REP(i, n)				for (int (i) = 0 ; (i) < (int)(n) ; ++(i))
#define REPN(i, m, n)			for (int (i) = m ; (i) < (int)(n) ; ++(i))
#define INF						2e9
#define MOD						(1000 * 1000 * 1000 + 7)
#define Ceil(x, n)				(((((x))+((n)-1))/n))		/* Nの倍数に切り上げ割り算 */
#define CeilN(x, n)				(((((x))+((n)-1))/n)*n)		/* Nの倍数に切り上げ */
#define FloorN(x, n)			((x)-(x)%(n))				/* Nの倍数に切り下げ */
#define IsOdd(x)				(((x)&0x01UL) == 0x01UL)			
#define IsEven(x)				(!IsOdd((x)))						
#define M_PI					3.14159265358979323846
typedef long long				ll;
typedef pair<ll, ll>			P;

/*-----------------------------------------------------------------------------
　処理
 -------------------------------------------------------------------------------*/
template<class T>
class UnionFind
{
private:
	vector<T> parentNo;
	vector<T> parentRank;
	vector<T> parentSize;

public:
	// コンストラクタ
	UnionFind(int n) 
	{
		parentNo.resize(n);
		parentRank.resize(n, 0);
		parentSize.resize(n, 1);
		for (int i = 0; i < n; i++) {
			// 始めは全部ROOT
			parentNo[i] = i;
		}
	}

	int root(T x) 
	{
		if (parentNo[x] == x) {
			return x;
		} else {
			T rootParentNo = root(parentNo[x]);
			parentNo[x] = rootParentNo;
			return rootParentNo;
		}
	}

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

	int size(T x) 
	{
		return parentSize[root(x)];
	}

	void unite(T xc, T yc) 
	{
		T xr = root(xc);
		T yr = root(yc);
		if (xr == yr) {
			return;
		}
		if (parentRank[xr] < parentRank[yr]) {
			parentNo[xr] = yr;
			parentSize[yr] += parentSize[xr];
		} else {
			parentNo[yr] = xr;
			parentSize[xr] += parentSize[yr];
			if (parentRank[xr] == parentRank[yr]) {
				parentRank[xr]++;
			}
		}
	}
};


int main()
{
	int N, M;
	cin >> N >> M;
	vector<int> p(N);
	REP(i, N) {
		cin >> p[i];
		p[i]--;
	};
	
	UnionFind<int> uFind(N);
	REP(i, M) {
		int x, y;
		cin >> x >> y;
		x--; y--;
		uFind.unite(x, y);
	}

	int ans = 0;
	REP(i, N) {
		if (uFind.same(p[i], i)) ans++;
	}
	cout << ans << endl; 
	return 0;
}