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<cstdio>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<bitset>
#include<string>
#include<stack>
#include<set>
#include<map>
#include<cstring>
#include<complex>
#include<cmath>
#include<iomanip>
#include<numeric>
#include<algorithm>
#include<list>
#include<functional>
#include<cassert>
#define mp make_pair
#define pb push_back
#define X first
#define Y second
#define INF 987654321987654321
#define PI 3.14159265358979323846264
#define fup(i,a,b,c) for(int (i)=(a);(i)<=(b);(i)+=(c))
#define fdn(i,a,b,c) for(int (i)=(a);(i)>=(b);(i)-=(c))
#define MEM0(a) memset(a,0,sizeof(a));
#define MEM_1(a) memset(a,-1,sizeof(a));
#define ALL(a) a.begin(),a.end()
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef vector<int> Vi;
typedef vector<ll> Vll;
typedef vector<double> Vd;
typedef vector<Pi> VPi;
typedef vector<Pll> VPll;
typedef vector<Pd> VPd;
typedef tuple<int, int, int> iii;
typedef tuple<ll, ll, ll> LLL;
typedef vector<iii> Viii;
typedef vector<LLL> VLLL;
typedef complex<double> base;
ll MOD = 1000000007;
ll POW(ll a, ll b, ll MMM = MOD) { ll ret = 1; for (; b; b >>= 1, a = (a*a) % MMM)if (b & 1)ret = (ret*a) % MMM; return ret; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { if (a == 0 || b == 0)return a + b; return a*(b / gcd(a, b)); }
int dx[] = { 0,-1,1,0 }, dy[] = { 1,0,0,-1 };
int ddx[] = { 0,0,1,1,1,-1,-1,-1 }, ddy[] = { 1,-1,1,0,-1,1,0,-1 };

int a[100001];
int parent[100001];
int col[100001], CN;
Vi v[100001];
int find(int a)
{
	if (parent[a] < 0)return a;
	return parent[a] = find(parent[a]);
}
void merge(int a, int b)
{
	a = find(a), b = find(b);
	if (a != b)
	{
		parent[a] += parent[b];
		parent[b] = a;
	}
}
int main() {
	MEM_1(parent);
	int n, m;
	scanf("%d%d", &n, &m);
	fup(i, 1, n, 1)scanf("%d", a + i);
	while (m--)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		merge(x, y);
	}
	fup(i, 1, n, 1)
		if (parent[i] < 0)col[i] = ++CN;
	fup(i, 1, n, 1)
	{
		if (parent[i] > 0)
			col[i] = col[find(i)];
		v[col[i]].push_back(i);
	}
	int ans = 0;
	fup(i, 1, CN, 1)
	{
		for (int x : v[i])
		{
			int t = a[x];
			int tt = lower_bound(ALL(v[i]), t) - v[i].begin();
			if (tt != (int)v[i].size() && v[i][tt] == t)ans++;
		}
	}
	printf("%d", ans);
}