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)

#define _CRT_SECURE_NO_WARNINGS

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

//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define FOR(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTL(d) printf("%lld\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9

typedef long long ll;
int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};
int par[100000];

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int a, int b) {
	return (a) > (b) ? (a) : (b);
}
void Swap(int *a, int *b) {
	int tmp = *a;
	*a = *b;
	*b = tmp;
}
int dfs(int i) {
	if (par[i] == i) {
		return i;
	}
	else {
		int t=dfs(par[i]);
		return par[i] = t;
	}
}

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	int p[100000];
	REP(i, n) {
		int t;
		scanf("%d", &t);
		p[i] = t - 1;
	}
	REP(i, n)par[i] = i;
	REP(i, m) {
		int x, y;
		scanf("%d %d", &x, &y);
		x--; y--;
		int xp = dfs(x);
		int yp = dfs(y);
		if (xp != yp) {
			par[yp] = xp;
		}
	}
	REP(i, n) {
		dfs(i);
	}
	/*
	REP(i, n) {
		printf("%d ",par[i]);
	}
	puts("");
	REP(i, n) {
		printf("%d ", p[i]);
	}
	puts("");
	*/
	int cnt = 0;
	REP(i, n) {
		if (par[p[i]] != par[i])cnt++;
		//printf("%d %d %d\n", par[p[i]], par[i], cnt);
	}
	PRINTD(n - cnt);
}