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'

n = gets.to_i
divisors = Prime.prime_division(n)

def f(a, b)
  [a.to_s.size, b.to_s.size].max
end

products = Array.new(divisors.size + 1) { Array.new }

products[0] << 1

divisors.each_with_index do |(divisor, number), index|
  (0..number).each do |num|
    products[index].each do |product|
      if num == 0
        products[index + 1] << product
      else
        products[index + 1] << product * divisor ** num
      end
    end
  end
end

ans = []
products.last.each do |a|
  b = n / a
  ans << f(a, b)
end
puts ans.min