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
primes = []
N.prime_division.each do |a|
    a[1].times do
        primes << a[0]
    end
end
ret = N
1.upto(primes.size) do |n|
    primes.combination(n).each do |ary|
        a = ary.inject(:*)
        b = primes.inject(:*).div(a) || 1
        ret = [ret, [Math.log10(a).to_i + 1, Math.log10(b).to_i + 1].max].min
    end
end
puts ret