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: Time Limit Exceeded)

require 'prime'
include Math
$n = gets.to_i
$prime = 2
$tmp1 = $n
$tmp2 = 1
$sq = sqrt($n)
=begin
R = RubyVM::InstructionSequence
R.compile_option = {tailcall_optimization: true, trace_instruction: false}
R.new(<<~EOS).eval

def culc
	if $n % $prime == 0
		$tmp1 = $n / $prime
		$tmp2 = $prime
		return $tmp2 if $tmp1 <= $tmp2
	end
	$prime += 1
	if $prime >=  $sq
		return $n
	else
		return culc
	end
end

EOS
=end
if Prime.prime?($n)
	puts log10($n).to_i+1
else
1.upto($sq * 250) do |prime|
  if $n % prime == 0
    $tmp1 = $n/prime
    $tmp2 = prime
    break if $tmp1 <= $tmp2
  end
end
	puts log10($tmp2).to_i+1
end