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>

using namespace std;

#define ll long long
#define ull unsigned long long
#define PI acos(-1.0)
#define eps 1e-12
#define fi first
#define se second
#define MEM(a,b) memset((a),(b),sizeof(a))
#define mod(x) ((x)%MOD)
#define pii pair<int,int>
#define wz cout<<"-----"<<endl;

const int maxn = 2e5 + 10, INF = 0x3f3f3f3f , MOD = 1e9 + 7;
const int Move[4][2] = {-1,0,1,0,0,1,0,-1};
const int Move_[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}


void init() {

}

int a[maxn];
vector<int> E[maxn];
int f[maxn];

int fnd(int x) {
    return x == f[x] ? x : f[x] = fnd(f[x]);
}

void sum(int x,int y)  {
    if (fnd(x) == fnd(y)) return ;
    f[fnd(x)] = fnd(y);
}

int m1[maxn],m2[maxn];
int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
//    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);
    ios_base::sync_with_stdio(0);
    init();
    int n, k;cin>>n>>k;
    for (int i = 0; i < maxn; i++) f[i] = i;
    for (int i = 1; i <= n; i++) cin>>a[i],m1[a[i]] = i;
    while (k--) {
        int x,y; cin>>x>>y;
        sum(x,y);
    }
    for (int i = 1; i <= n; i++) {
        sum(i,f[i]);
    }
    for (int i = 1; i <= n; i++) {
        //cout<<i<<' '<<f[i]<<endl;
        E[f[i]].push_back(i);
    }
    int res = 0;
    for (int i = 1; i <= n; i++) {
        if(f[i] != i) {
            //res++;
        } else {
            //res--;
            for (int j = 0; j < E[i].size(); j++){
               // cout<<E[i][j]<<' ';
                m2[E[i][j]] = i;
            }
            //cout<<endl;
            for (int j = 0; j < E[i].size(); j++){
                if (m2[a[E[i][j]]] == i) res++;
            }
        }
    }
    cout<<res<<endl;
    return 0;
}