Problem:
Score : 400 points

Problem Statement

There are N empty boxes arranged in a row from left to right.
The integer i is written on the i-th box from the left (1 \leq i \leq N).

For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.

We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:

For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.

Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

a_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If a good set of choices does not exist, print -1.

If a good set of choices exists, print one such set of choices in the following format:

M
b_1 b_2 ... b_M

where M denotes the number of boxes that will contain a ball, and b_1,\ b_2,\ ...,\ b_M are the integers written on these boxes, in any order.

Sample Input 1

3
1 0 0

Sample Output 1

1
1

Consider putting a ball only in the box with 1 written on it.

There are three boxes with multiples of 1 written on them: the boxes with 1, 2, and 3. The total number of balls contained in these boxes is 1.

There is only one box with a multiple of 2 written on it: the box with 2. The total number of balls contained in these boxes is 0.

There is only one box with a multiple of 3 written on it: the box with 3. The total number of balls contained in these boxes is 0.

Thus, the condition is satisfied, so this set of choices is good.

Sample Input 2

5
0 0 0 0 0

Sample Output 2

0

Putting nothing in the boxes can be a good set of choices.

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

const nmax=200010;
type ary=array[1..nmax] of 0..1;
var N,i,j,k,latter,sum,count:int64;
    a,ans:ary;
begin
read(N);
for i := 1 to N do 
 begin read(a[i]);
  if i > N div 2 then 
   begin 
    ans[i]:=a[i];inc(latter)
   end
  end;
for i := N-latter downto 1 do
 begin
  sum:=0;
  for j := 1 to N do
   if i*j > N then break 
   else sum:=sum+a[i*j];
  ans[i]:=sum mod 2;
  if ans[i]=1 then inc(count)
 end;
writeln(count);
for i := 1 to N do
 if ans[i]=1 then write(i,' ');
end.