Problem: Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

Solution (in PYTHON3):

Solution:
zeroes = int(input().strip())
two, five = 0, 0

def counttwofive(n):
	t, f = 0, 0
	while n % 2 == 0:
		t = t + 1
		n = n / 2
	while n % 5 == 0:
		f = f + 1
		n = n / 5
	return (t, f)

i = 1
ret = []

while min(two, five) <= zeroes:
	c = counttwofive(i)
	two = two+c[0]
	five = five+c[1]
	
	if min(two, five) == zeroes:
		ret.append(i)
	
	i = i + 1

print(len(ret))
print(" ".join(str(s) for s in ret))