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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
#include <iomanip>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<long double> VD;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef vector<VD> VVD;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }
#define INF 1LL<<60

void dfs1(int from, vector< vector<int> >& edges, vector<int>& group, int group_idx) {
    if(group[from] > 0) return;
    group[from] = group_idx;

    for(int next: edges[from]) {
        dfs1(next, edges, group, group_idx);
    }

    return;
}

int main() {
    int N, M;
    cin >> N >> M;
    vector<int> p(N), group(N);
    vector<bool> visited(N);
    vector< vector<int> > edges(N, vector<int>());
    vector< set<int> >reachable(N, set<int>());
    REP(i, N) {
        cin >> p[i];
        p[i]--;
    }
    REP(i, M) {
        int tmp1, tmp2;
        cin >> tmp1 >> tmp2;
        edges[tmp1-1].push_back(tmp2-1);
        edges[tmp2-1].push_back(tmp1-1);
    }

    int group_idx = 1;
    REP(i, N) {
        if(group[i] == 0) {
            dfs1(i, edges, group, group_idx);
            group_idx++;
        }
    }

    int ans = 0;
    REP(i, N) {
        if(group[i] == group[p[i]]) ans++;
    }

    cout << ans << endl;

    return 0;
}