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'
N = gets.to_i

divs = N.prime_division.flat_map{|d, n|
  [d] * n
}
divs = [1] if divs.empty?
len = divs.length
ans = Float::INFINITY
for b in 0..(2**len-1)
  pat = format("%0#{len}b", b).chars
  u = v = 1
  pat.zip(divs) do |group, d|
    group == "0" ? u *= d : v *= d
  end
  uu = u.to_s.length
  vv = v.to_s.length
  ans = [ans, [uu, vv].max].min
end
p ans