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: Time Limit Exceeded)

#include<stdio.h>

struct num
{
    int n1;
    int n2;
};

int ifpass(int s1,int s2,struct num narr[],int m,int n);

int main()
{
    int n,m,i,j,ans=0;
    scanf("%d %d",&n,&m);
    int a[100000];
    for(i=0;i<n;i++)
    {
        scanf("%d",&(a[i]));
    }
    struct num narr[100000]={0};
    int p1,p2;
    for(i=0;i<m;i++)
    {
        scanf("%d %d",&p1,&p2);
        if(p1<p2)narr[i].n1=p1,narr[i].n2=p2;
        else narr[i].n1=p2,narr[i].n2=p1;
    }
    for(i=1;i<=n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i==a[j])
            {	//printf("\n%d to %d then %d",i,j+1,ifpass(i,(j+1),narr,m,n));
                ans+=ifpass(i,(j+1),narr,m,n);
            }
        }
    }
	printf("%d",ans);
}

int ifpass(int s1,int s2,struct num narr[],int m,int n)
{
	//printf("pass(%d %d)",s1,s2);
    if(s1==s2)return 1;
    int k,k1,k2,flag=0;
	int yarr[100001]={0};
	yarr[s1]=1;//キャリア
	while(!flag)
	{
		flag=1;
		for(k=0;k<m;k++)
		{
			if(yarr[narr[k].n1]==1)
			{
				if(yarr[narr[k].n2]!=1)
				{
					yarr[narr[k].n2]=1;
					flag=0;
				}
			}
			else if(yarr[narr[k].n2]==1)
			{
				if(yarr[narr[k].n1]!=1)
				{
					yarr[narr[k].n1]=1;
					flag=0;
				}
			}
		}
	}
	
    return yarr[s2];
}