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)

#!/usr/bin/env ruby
# require 'pry'

# N
# a_1 b_1
# ...
# a_N b_N
# ↑を[[a_1,..a_N],[b_1,..b_N]]にする
# N = gets.to_i
# a, b = N.times.map{gets.split.map(&:to_i)}.transpose

n = gets.to_i
root_n = Integer.sqrt(n)
min_f = n.to_s.length
1.upto(root_n) do |i|
  if n % i == 0
    min_f = (n/i).to_s.length
  end
end

puts min_f