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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

#define REP(i, N) for (int i = 0; i < (int)(N); i++)

// chmax, chmin
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const int MAX_N = 1e+5 + 10;

int par[MAX_N]; // 親
int Rank[MAX_N]; // 木の深さ

// n要素で初期化
void init(int n) {
    for (int i = 0; i < n; i++) {
        par[i] = i;
        Rank[i] = 0;
    }
}
// 木の根を求める
int find(int x) {
    if (par[x] == x) {
    return x;
    } else {
    return par[x] = find(par[x]);
    }
}
// xとyの属する集合を併合
void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (Rank[x] < Rank[y]) {
        par[x] = y;
    } else {
        par[y] = x;
    if (Rank[x] == Rank[y]) Rank[x]++;
    }
}
// xとyが同じ集合に属するか否か
bool same(int x, int y) {
    return find(x) == find(y);
}

int main() {
    int N,M;
    cin >> N >> M;
    vector<int> p(N),x(M),y(M);
    REP(i,N) {
        int c;
        cin >> c;
        --c;
        p[i] = c;
    }
    REP(i,M) {
        int a,b;
        cin >> a >> b;
        --a; --b;
        x[i] = a;
        y[i] = b;
    }
    
    init(N);
    REP(i,M){
        unite(p[x[i]],p[y[i]]);
    }
    
    int ans = 0;
    REP(i,N){
        if ( same(i,p[i]) ) ++ans;
    }
    
    cout << ans << endl;
    
    
    return 0;
}