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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Data.List
import System.Environment (getArgs)

-- http://tnomura9.exblog.jp/12332000/
factors :: Integer -> [Integer]
factors n = [x | x <- [1..n], n `mod` x == 0]

factorization :: Integer -> [Integer]
factorization 1 = []
factorization x = v : factorization (x `div` v)
  where v = (factors x) !! 1

factorList n p (x:xs) | (p*x*p*x) >= n = [p] ++ (factorList n p xs)
                      | otherwise = [p*x] ++ (factorList n (p*x) xs)
factorList n p [] = [p]

answer n | n < 10 = 1
         | otherwise = minimum $ map f $ nub $ sort $ factorList n (head l) (drop 1 l)
  where f i = maximum $ map g [i, n `div` i]
        g x = length (show x)
        l = reverse $ sort $ factorization n

main = do
  n <- readLn
  print $ answer (n :: Integer)