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: Wrong Answer)

#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <iterator>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

constexpr int MAX_N = 100000;

int N, M;
int p[MAX_N];
std::map<int, int> idx;
std::vector<std::vector<int> > vss;
int ans;

int main()
{
  scanf( "%d%d", &N, &M );
  rep( i, N )
    scanf( "%d", p+i ), --p[i], ans += p[i]==i;

  rep( i, M )
  {
    int x, y;
    scanf( "%d%d", &x, &y );
    --x; --y;

    int id = -1, oth;
    if( idx.count( x ) && idx.count( y ) )
    {
      int idX = idx[x], idY = idx[y];

      rep( j, vss[idY].size() )
        idx[vss[idY][j]] = idX;

      std::copy( all(vss[idY]), std::back_inserter(vss[idX]) );
      vss.erase( vss.begin()+idY );
    }
    else if( idx.count( x ) )
      id = idx[x], oth = y;
    else if( idx.count( y ) )
      id = idx[y], oth = x;
    
    if( id == -1 )
    {
      vss.push_back( std::vector<int>( {x, y} ) );
      idx[x] = idx[y] = vss.size()-1;
    }
    else
      vss[id].push_back( oth ); 
  }

  rep( i, vss.size() )
  {
    std::map<int, int> cnt;

    rep( j, vss[i].size() )
      ++cnt[p[vss[i][j]]];
      
    rep( j, vss[i].size() ) if( p[vss[i][j]] != vss[i][j] )
      ans += cnt.count( vss[i][j] );
  }
  
  printf( "%d\n", ans );

  return 0;
}