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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ull=unsigned long long;
#define mod 1000000007
#define mod2 998244353 
#define PI 3.14159265
#define ll long long
#define ld long double
#define pi pair<ll,ll>
#define pb push_back
#define vi vector<ll>
#define sz size()
#define setbits(x)  __builtin_popcount(x)
#define fi first
#define se second
#define lz length()
#define all(x) (x).begin(),(x).end()
#define scf(x) scanf("%lld",&x)
#define rep(i,n) for(ll i=0;i<n;i++)
const int INF = (int)1e9;
const ll IINF=1LL<<62;
const int maxn=10000005;

//int divr[maxn];
/**/
/*void divisor()
{for(int i=2;i<sqrt(maxn);i++)
     {
          if(!divr[i])
          {
               for(int j=i*i;j<maxn;j+=i)if(!divr[j])divr[j]=i;     
          }
     }
     for(int i=1;i<maxn;i++)if(!divr[i])divr[i]=i;
}*/
 
/*const int N=1000005;
ll n;
struct line {
    long long m, c;
    long long eval(long long x) { return m * x + c; }
    long double intersectX(line l) { return (long double) (c - l.c) / (l.m - m); }
};
vector<pair<pi,ll>> rect;
deque<line> dq;
*/
ll modexpo(ll a,ll b)
{
     ll res=1LL;
     while(b>0)
     {
          if(b&1)res=(res*a)%mod;
          a=(a*a)%mod;
          b>>=1;
     }
     return res;
}
 
 
 
 
 
 
 







const int maxn1=100005;
vector<int> adj[maxn1];
int p[maxn1];
bool visited[maxn1];
vector<int> chain,value;
int ans=0;
void dfs(int node){
     visited[node]=true;
  	 //cout<<node<<" "<<p[node]<<"\n";
  	 chain.push_back(node);value.push_back(p[node]);
     for(int child:adj[node]){
          if(!visited[child])
          dfs(child);
     }

}
int buf[maxn1];
int main()
{
     int n,m;
     cin>>n>>m;
     for(int i=1;i<=n;i++)cin>>p[i];
     for(int i=1;i<=m;i++){
          int x,y;
          cin>>x>>y;
          adj[x].pb(y);
          adj[y].pb(x);
     }
     for(int i=1;i<=n;i++){
          if(!visited[i]){
            dfs(i);
               for(auto x:chain)buf[x]=1;
               for(auto x:value)ans+=buf[x];
               for(auto x:chain)buf[x]=0;
            //cout<<ans<<"\n";
               chain.clear();value.clear();
          }

     }
     cout<<ans;



}