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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include<queue>
using namespace std;
//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
//math
//-------------------------------------------
template<class T> inline T sqr(T x) {return x*x;}
//typedef
//------------------------------------------
typedef long long ll;
typedef long long LL;
typedef vector<int> VI;
typedef vector<long long > VLL;
typedef vector<long long > vll;
typedef pair <long long , long long > pll;
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define VECMAX(x) *max_element(ALL(x))
#define VECMIN(x) *min_element(ALL(x))
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define MULTIPLE(i,n,k) for(int i = (k) ; i<(n) ; i+=k+1)//倍数ループ
//constant
//------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
#define SIZEOF(x) sizeof(x)/sizeof(x[0])




//-------------------DIVIDE----------------------
// DIV[i][j] は　i の j分割数 j == 0 && i != 0 なら 0  



//----UnionFind-----

class UnionFind{
    public:
    VI par;
    UnionFind(LL N):par(N){
        REP(i,N)par[i] = i;
    }
    ~UnionFind(){}
    LL root(LL x){
        if(par[x] ==x)return x;
        else {
            par[x] = root(par[x]);
            return par[x];
        }
    }
    void unite(LL x, LL y){
        LL rx = root(x);
        LL ry = root(y);
        if(rx == ry)return;
        par[rx] = ry;
    }
    bool same(LL x, LL y){
        LL rx = root(x);
        LL ry = root(y);
        return rx == ry;
    }
};




//-----------MAIN------------//




int main(){
    LL n,m;
    cin >> n >> m;
    vll p(n+1);
    REP(i,n){
        cin >> p[i];

        p[i]--;

    }
    map<ll , ll> Isgrouped;
    UnionFind U(n);
    REP(i,m){
        ll aaa;
        ll bbb;
        cin >> aaa >> bbb;
        U.unite(aaa-1,bbb-1);
        Isgrouped[aaa-1] = 1;
        Isgrouped[bbb-1] = 1;
    
    }   
    vector<map <ll , ll> >RootGroupe(n+1);
    REP(i,n){
        RootGroupe[U.root(i)][i] = 1;
    }
    ll count = 0;
    REP(i,n){
        if(Isgrouped[p[i]] == 1){
            if(RootGroupe[U.root(p[i])][i] == 1){
                count += 1;
            }
        }else{
            if(p[i] == i)count += 1;
        }
    }


    cout << count;






    
    return 0;

}