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 PYTHON):

Solution:
#! /usr/bin/python

# A simple solution

def two_and_fives(x):
    two = 0
    five = 0
    while x % 2 == 0:
        two += 1
        x /= 2
    while x % 5 == 0:
        five += 1
        x /= 5
    return [two, five]

m = int(raw_input())
num_2_5 = [0, 0]
r = []
i = 1
while True:
    [two, five] = two_and_fives(i)
    num_2_5[0] += two
    num_2_5[1] += five
    zeros = min(num_2_5)
    if zeros == m:
        r.append(i)
    elif zeros > m:
        break
    i += 1
print(len(r))
print(" ".join([str(e) for e in r]))