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 <stdio.h>
#include <string.h>
#include <iostream>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <algorithm>
#include <numeric>
#include <utility>
#include <cmath>

using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;

const int iinf = 1 << 29;
const int imax = ~(1 << 31);
const long long linf = 1ll << 61;
const long long lmax = ~(1ll << 63);

#define dump(x) cerr << #x << " : " << x << '\n'
#define MOD(x, y) (((y) + (x) % (y)) % (y))

int par[200000];
int level[200000];

void init(int n)
{
    for (int i = 0; i < n; i++) {
        par[i] = i;
        level[i] = 0;
    }
}

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 (level[x] < level[y]) {
        par[x] = y;
    } else {
        par[y] = x;
        if(level[x] == level[y]) level[x]++;
    }
}

bool same(int x, int y)
{
    return find(x) == find(y);
}

int N, M;
int pos[200000];

int main(int argc, char* argv[])
{
    scanf("%d%d", &N, &M);
    init(N);
    for (int i = 0; i < N; i++) {
        int p;
        scanf("%d", &p);
        p--;
        pos[p] = i;
    }
    for (int i = 0; i < M; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        x--; y--;
        unite(x, y);
    }
    int ans = 0;
    for (int i = 0; i < N; i++) {
        if (same(pos[i], i)) {
            ans++;
        }
    }
    printf("%d\n", ans);

	return 0;
}