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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using static System.Console;
using static System.Math;

namespace CP
{
    class Atria
    {
        readonly static long mod = 1000000000 + 7;
        static void Main(string[] args)
        {
            long n = long.Parse(ReadLine());
            var ls = PrimeFactor(n).ToList();
            var tmp = n;
            var ans = F(1, tmp);
            for (int i = 0; i < ls.Count; i++)
            {
                tmp /= ls[i];
                var tmp2 = n / tmp;
                ans = Min(ans, F(tmp2, tmp));
            }
            WriteLine(ans);
        }

        static int F(long a, long b)
        {
            return Max(Digits(a), Digits(b));
        }

        static int Digits(long a)
        {
            return(a == 0) ? 1: ((int)Log10(a) + 1);
        }
        static IEnumerable<long> PrimeFactor(long n)
        {
            int i = 2;
            long tmp = n;

            while (i <= Math.Sqrt(n))
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    yield return i;
                }
                else i++;
            }

        }

    }
}