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 (2.7.1) (judge verdict: Accepted)

require 'prime'

n = gets.to_i
if n == 1
  puts 1
else
  h = Prime.prime_division(n).to_h
  p = []
  h.each do |k, v|
    while v > 0
      p << k
      v -= 1
    end
  end
  min = Float::INFINITY
  0.upto(2 ** p.size - 1) do |bit|
    a = 1
    b = 1
    p.size.times do |i|
      if bit[i] == 0
        a *= p[i]
      else
        b *= p[i]
      end
    end
    min = [a.to_s.size, b.to_s.size].max if min > [a.to_s.size, b.to_s.size].max
  end
  puts min
end