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:
def test():
    from math import factorial as f
    cnt = 0
    i = 1
   #while cnt != 100000+1: 
    for i in range(1, 100+1):
        s = str(f(i))
        cnt = 0
        for j in range(len(s)-1,-1,-1):
            if s[j] == "0":
                cnt += 1
            else:
                break
        print(cnt, end= " ")#, f(i))
        i += 1
    print(cnt, end=" ")
    


def func(x):
    ans = 0
    while x % 5 == 0:
        x //= 5
        ans += 1
    return ans

def main():
    n = int(input())
    cnt = 0
    i = 1
    while True:
        if i % 5 == 0:
            cnt += func(i)
        if cnt == n:
            print(5)
            print(i, i+1, i+2, i+3, i+4)
            break
        if cnt > n:
            print(0)
            break
        i += 1

main()