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)

require 'prime'
N = gets.chomp.to_i
a = N.prime_division
divisors =[1] # 約数
a.each do |ary|
  ary[1].times do |n|
    divisors << ary[0]
  end
end
F_all = []
1.upto(divisors.size()-1) do |num|
  ones = divisors.combination(num).to_a.uniq()
  ones.each do |one|
    other = divisors.dup
    one.each do |data|
      other.delete_at(other.index(data))
    end
#    print("片方：#{one}, 残り：#{other}\n")
#    print("片方桁数: #{one.inject(:*).to_s.split("").size()}, ")
#    print("残り桁数: #{other.inject(:*).to_s.split("").size()}\n\n")
    F_all << [one.inject(:*).to_s.split("").size(), other.inject(:*).to_s.split("").size()].max
  end
end
print("#{F_all.min}")