Problem:
Score : 300 points

Problem Statement

You are given an integer N.

For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.

For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.

Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

1 \leq N \leq 10^{10}

N is an integer.

Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

n=gets.to_i


def factor(num)
  result = []
  i = 2
  while i <= num do
    if num % i == 0
      result << i
      num = num / i
    else
      i += 1
    end
  end
  return result
end
a=factor(n)

diva=1
count=[]
for i in 0..a.length-1
diva=diva*a[i]
b=n.div(diva)

diva_s=diva.to_s
b_s=b.to_s
if diva_s.length>b_s.length
    count.push(diva_s.length)
else
    count.push(b_s.length)
end
end
p a.inject(&:*)
puts count.min