Problem: Now you get Baby Ehab's first words: "Given an integer n, find the longest subsequence of [1,2, …, n-1] whose product is 1 modulo n." Please solve the problem.

A sequence b is a subsequence of an array a if b can be obtained from a by deleting some (possibly all) elements. The product of an empty subsequence is equal to 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

The first line should contain a single integer, the length of the longest subsequence.

The second line should contain the elements of the subsequence, in increasing order.

If there are multiple solutions, you can print any.

Examples

Input


5


Output


3
1 2 3 

Input


8


Output


4
1 3 5 7 

Note

In the first example, the product of the elements is 6 which is congruent to 1 modulo 5. The only longer subsequence is [1,2,3,4]. Its product is 24 which is congruent to 4 modulo 5. Hence, the answer is [1,2,3].

Solution (in PYTHON3):

Solution:
import math
  
# A function to print all prime factors of 
# a given number n
def primeFactors(n):
      
    # Print the number of two's that divide n
    while n % 2 == 0:
        prim.append(2),
        n = n / 2
          
    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3, int(math.sqrt(n))+1, 2):
          
        # while i divides n, print i ad divide n
        while n % i == 0:
            prim.append(int(i)),
            n = n / i
              
    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        prim.append(int(n))
prim=[]
n=int(input())
ans=[1 for i in range(n+1)]
primeFactors(n)

prim=list(set(prim))
prim.sort()
for i in prim:
    for j in range(i,n+1,i):
       
        ans[j]=0
an=1
count=0
arr=[]
for i in range(1,n-1):
    if(ans[i]==1):
        count+=1
        an=(an*i)%n
        arr.append(i)
if((an*(n-1))%n==1):
    an=(an*(n-1))%n
    count+=1
    arr.append(n-1)
if(an!=1):
    print(-1)
print(count)
print(*arr)