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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

typedef struct node node;
typedef struct edge{
    node* target;
    struct edge* next;
}edge;
typedef struct node{
    int id;
    edge* fst;
    edge* lst;
}node;
void initNodeArray(node* p, int n){
    for(int i=0; i<n; i++){
        (p+i)->id = 0;
        (p+i)->fst = NULL;
        (p+i)->lst = NULL;
    }
}
void addEdge(node* p, node* target){
    if(p->fst == NULL){
        p->fst = (edge*)malloc(sizeof(edge));
        p->lst = p->fst;
    }else{
        p->lst->next = (edge*)malloc(sizeof(edge));
        p->lst = p->lst->next;
    }
    p->lst->target = target;
    p->lst->next = NULL;
}
void rmEdge(node* p, node* q){
    edge* e = p->fst;
    edge* pre = NULL;
    while(e != NULL){
        if(e->target == q){
            if(pre != NULL){
                pre->next = e->next;
            }
            if(e == p->fst){
                p->fst = e->next;
            }
            if(e == p->lst){
                p->lst = pre;
            }
            free(e);
            return;
        }
        pre = e;
        e = e->next;
    }
    return;
}

void func(int* group, node* a, int g){
    if(*(group + (a->id)) != 0){
        return;
    }
    *(group + (a->id)) = g;
    edge* e = a->fst;
    while(e != NULL){
        func(group, e->target, g);
        e = e->next;
    }
}

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int p[n], q[n];
    node a[n];
    initNodeArray(a, n);
    int group[n];
    rep(i, 0, n){
        scanf("%d", &p[i]);
        a[i].id = i;
        group[i] = 0;
    }
    rep(i, 0, m){
        int x, y;
        scanf("%d %d", &x, &y);
        x--; y--;
        addEdge(&a[x], &a[y]);
        addEdge(&a[y], &a[x]);
    }
    int num = 1;
    rep(i, 0, n){
        func(group, &a[i], num++);
    }
    rep(i, 0, n){
        q[p[i]-1] = group[i];
    }
    int ans = 0;
    rep(i, 0, n){
        if(group[i] == q[i]){
            ans++;
        }
    }
    printf("%d", ans);

    return 0;
}