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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var i,j:longint;
    fx,fy,x,y:longint;
    n,m:longint;
    a,next,last,father:array[0..200050]of longint;
    v,newv:array[0..200050]of boolean;
    ans:longint;
function gf(x:longint):longint;
begin
  if father[x]=x then exit(x) else
  begin
    father[x]:=gf(father[x]);
    gf:=father[x];
  end;
end;
begin
  read(n,m);
  for i:=1 to n do
  read(a[i]);
  for i:=1 to n do
  father[i]:=i;
  for i:=1 to m do
  begin
    read(x,y);
    fx:=gf(x);
    fy:=gf(y);
    if fx<>fy then father[fx]:=fy;
  end;
  for i:=n downto 1 do
  begin
    next[i]:=last[gf(i)];
    last[gf(i)]:=i;
  end;
  for i:=1 to n do
  if not v[i] then
  begin
    x:=i;
    repeat
      newv[x]:=true;
      x:=next[x];
    until x=0;
    x:=i;
    repeat
      if newv[a[x]] then inc(ans);
      x:=next[x];
    until x=0;
    x:=i;
    repeat
      v[x]:=true;
      newv[x]:=false;
      x:=next[x];
    until x=0;
  end;
  writeln(ans);
end.