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: Accepted)

require 'prime'
n = STDIN.gets.to_i
divisors = [1]
primes = []
n.prime_division.each do |prime|
  prime[1].times { primes << prime[0] }
end

1.upto(primes.size) do |i|
  primes.combination(i) do |prime|
    divisors << prime.reduce(:*)
  end
end

divisors = divisors.uniq.sort
min = n.to_s.size
divisors.uniq.sort.each do |num|
  o = n / num
  m = [o.to_s.size, num.to_s.size].max
  if m < min
    min = m
  end
end
puts min