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: Accepted)

require 'prime'

def common_divisors(n)
  return [1] if n == 1

  tmp = n.prime_division.map do |e|
    (0..e[1]).to_a.map do |i|
      e[0]**i
    end
  end
    
  tmp.reduce{|p,q| p.product(q)}.map do |a|
    [a].flatten.reduce(&:*)
  end.sort
end

n = gets.to_i

div1 = common_divisors(n)
div2 = div1.reverse

ans = Float::INFINITY
div1.size.times do |i|
  digits1 = div1[i].to_s.length
  digits2 = div2[i].to_s.length
  f = [digits1, digits2].max
  ans = [ans, f].min
end

puts ans