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<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a=c;}
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod(ll n){return n%=MOD1;}
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct{ ll aa , bb;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->aa - ((fr*)q)->aa;}

/*-------------------------------------------------*/
// Union Find              添字に注意!!!!!!!!!!
typedef struct{
    ll num; //number
    ll par; //parent
    ll height; //height
}UF;
void UFmake(ll n, UF b[n]){for(ll i=0;i<n;i++){
    b[i].par = i;
    b[i].num = i;
    b[i].height = 1;
    }
}
ll UFroot(ll n, ll a,UF b[n]){ return a==b[a].par?a:UFroot(n,b[a].par,b); }
void UFunite(ll n, ll x, ll y, UF b[n]){
    if(b[UFroot(n,x,b)].height > b[UFroot(n,y,b)].height) b[UFroot(n,y,b)].par = UFroot(n,x,b);
    else{ b[UFroot(n,x,b)].par = UFroot(n,y,b); b[UFroot(n,y,b)].height = max2(b[UFroot(n,y,b)].height,b[UFroot(n,x,b)].height+1);}
}
ll UFsame(ll n, ll x, ll y, UF b[n]){ return UFroot(n,x,b) == UFroot(n,y,b) ? 1 : 0; }

int main(void){
    // char ;
    ll n,m,ans=0;
    scanf("%lld%lld",&n,&m);
    ll p[n],x[m],y[m];
    rep(i,0,n){
        scanf("%lld",&p[i]);
    }
    rep(i,0,m){
        scanf("%lld%lld",&x[i],&y[i]);
    }
    UF tr[n];
    UFmake(n,tr);
    rep(i,0,m){
        UFunite(n,x[i]-1,y[i]-1,tr);
    }
    rep(i,0,n){
        if(UFsame(n,p[i]-1,p[p[i]-1]-1,tr)){
            ans++;
        }
    }
    printf("%lld\n",ans);
    return 0;
}