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: Wrong Answer)

#include"bits/stdc++.h"
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define INF 1e+9


int main() {
	int n, m;
	cin >> n >> m;

	char p[100001];
	for (int i = 1; i <= n; i++)
	{
		cin >> p[i];
	}

	map<int, string> mp;
	int i=0;
	char x, y;
	cin >> x >> y;
	mp[0] = x + y;
	for (int j = 1; j < m; j++)
	{
		cin >> x >> y;
		if (mp[i].find(x) == true)
			if (mp[i].find(y) == false)
				mp[i] += y;
		if (mp[i].find(y) == true)
			if (mp[i].find(x) == false)
				mp[i] += x;
		if (mp[i].find(x) == false)
			if (mp[i].find(y) == false)
			{
				i++;
				mp[i] = x + y;
			}
	}

	int ans = 0;
	map<int, string>::iterator it;
	for (int j = 1; j <= n; j++)
	{
		for (it = mp.begin(); it != mp.end(); it++)
		{
			pair<int, string> pair = *it;
			if (pair.second.find(p[j]) != false)
			{
				char k = (char)j;
				if (pair.second.find(j) != false)
					ans++;
			}
		}
	}

	cout << ans << endl;

	return 0;
}