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>
#define int long long
#define mod (int)(1e9+7)
#define inf (int)(3e18+7)
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define P pair<int,int>
#define all(v) v.begin(),v.end()
#define mkp make_pair
#define mkt make_tuple
#define prique(T) priority_queue<T,vector<T>,greater<T>>
#define vecunique(vec) sort(vec.begin(), vec.end());decltype(vec)::iterator result = std::unique(vec.begin(), vec.end());vec.erase(result, vec.end())
using namespace std;

bool prime(int x) {
	for (int i = 2; i * i <= x; i++) {
		if (x % i == 0)return false;
	}
	return x > 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x * y / gcd(x, y);
}
int kai(int x) {
	if (x == 0)return 1;
	return kai(x - 1) * x % mod;
}
int mod_pow(int x, int y, int mod_) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % mod_;
		}
		x = x * x % mod_;
		y >>= 1;
	}
	return res;
}
int comb(int x, int y) {
	return kai(x)* mod_pow(kai(x - y), mod - 2, mod) % mod * mod_pow(kai(y), mod - 2, mod) % mod;
}
/*--------Library Zone!--------*/

struct union_find {
	int par[214514], size_[214514];
	union_find(int x) {
		rep(i, x) { par[i] = i; size_[i] = 1; }
	}
	int find(int x) {
		if (par[x] == x)return x;
		return par[x] = find(par[x]);
	}
	void unite(int x, int y) {
		x = find(x); y = find(y);
		if (x == y)return;
		if (size_[x] < size_[y]) {
			par[x] = y;
			size_[y] += size_[x];
		}
		else {
			par[y] = x;
			size_[x] += size_[y];
		}
	}
};

int n, m;
int a[114514];
signed main() {
	cin >> n >> m;
	REP(i, n + 1)cin >> a[i];
	union_find uf(n + 1);
	rep(i, m) {
		int b, c; cin >> b >> c; uf.unite(b, c);
	}
	int ans = 0;
	REP(i, n + 1) {
		if (uf.find(i) == uf.find(a[i]))ans++;
	}
	cout << ans << endl;
}