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'

class Integer
  def my_divisor_list
    divisors = [1]
    primes = []
    Prime.prime_division(self).each do |prime|
      prime[1].times {primes << prime[0]}
    end

    1.upto(primes.size) do |i|
      primes.combination(i) do |prime|
        divisors << prime.inject{|a,b| a *= b}
      end
    end

    divisors.uniq!
    divisors.sort!
    return divisors
  rescue ZeroDivisionError
    return
  end
end

N = gets.to_i
divisiors = N.my_divisor_list.select {|x| x <= Math.sqrt(N)}

ans = Float::INFINITY
f = 0

divisiors.each do |a|
	f = [a.to_s.length, (N/a).to_s.length].max
	ans = f if f < ans
end

p f