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:
import io, os
import sys
from math import log
from atexit import register

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.stdout = io.BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))

tokens = []
tokens_next = 0


def nextStr():
    global tokens, tokens_next
    while tokens_next >= len(tokens):
        tokens = input().split()
        tokens_next = 0
    tokens_next += 1
    return tokens[tokens_next - 1]


def nextInt():
    return int(nextStr())


def nextIntArr(n):
    return [nextInt() for i in range(n)]


def print(s, end='\n'):
    sys.stdout.write((str(s) + end).encode())


def countFivesInFac(n):
    res = 0
    i = 5
    while n / i >= 1:
        res += n // i
        i *= 5
    return res


def solve(m):
    low = 1
    high = 10**6
    while low + 1 < high:
        mid = (low + high) // 2
        if countFivesInFac(mid) < m:
            low = mid
        else:
            high = mid

    return list(range(high, high + 5)) if countFivesInFac(high) == m else []


if __name__ == "__main__":
    m = nextInt()
    res = solve(m)
    print(len(res))
    print(' '.join(map(str, res)))