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:
# найти k чисел, которые делятся на 10^m
#10 factors 5 and 2
#5, 10, 15, 20, 25, ..., 50, ..., 75, ..., 100, 125
# 5 - 20 one zero
# 25 - 100 two zero 125 - 625 three zero floor(5throot)
# 4(1+2+3+...) =
# 1 2 3 4 6 7 8 9 10 12 15 18 21 24
# 5 10 15 20 25 30 ...
import math
def toAdd(n):
    cnt = 0
    while(n != 1):
        if(n % 5 == 0):
            cnt += 1
            n //= 5
        else:
            break
    return cnt

cnt = 0
exist = []

curr = 5
while cnt <= 100000:
    cnt += toAdd(curr)
    exist.append(cnt)
    curr += 5


m = int(input())
if m in exist:
    print(5)
    for i in range(5 * (exist.index(m) + 1), 5 * (exist.index(m) + 2)):
        print(i, end = " ")
else:
    print(0)