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: Wrong Answer)

n=gets.to_i

data=[1]
(2..Math.sqrt(n).ceil).each do |i|
    while n % i == 0
        n /= i
        data << i
    end
end
if n != 1
    data << n
end

data.sort!.reverse!
if data.size == 1
    puts 1
else
    l, r = data.shift, data.shift
    data.each do |n|
        if l > r
            r *= n
        else
            l *= n
        end
    end
    l, r = l.to_s.length, r.to_s.length
    puts l > r ? l : r
end