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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp4
{
    class Program
    {
        static void Main(string[] args)
        {
            long n = long.Parse(Console.ReadLine());
            Console.WriteLine(Solve(n));
        }

        static int Solve(long n)
        {
            long[] divisors = Divisors(n);

            if (divisors.Length % 2 == 1)
            {
                return DigitsCount(divisors[divisors.Length / 2]);
            }
            else
            {
                return Math.Max(DigitsCount(divisors[divisors.Length / 2]), DigitsCount(divisors.Length / 2 - 1));
            }
        }

        static long[] Divisors(long n)
        {
            long sqrt = (long)Math.Ceiling(Math.Sqrt(n));
            var list = new List<long>();

            for (long i = 1; i <= sqrt; i++)
            {
                if (n % i == 0)
                {
                    list.Add(i);
                    list.Add(n / i);
                }
            }

            list.Sort();
            return list.ToArray();
        }

        static int DigitsCount(long n) => (int)Math.Log10(n) + 1;
    }
}