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: Time Limit Exceeded)

import Data.Char
import Data.List
main::IO()
main = do
    n <- readLn :: IO Double
    let r =  (floor . fromRational . toRational . sqrt) n
        m = truncate n
        ans = if (isP . fromIntegral) m then digitLen m
                                        else digitLen $ solve m r

    print $ ans


solve::Integer -> Integer-> Integer 
solve m r = case find (\x -> m `mod` x == 0) [r,(r + 1)..m] of
                 Just x -> x
                 Nothing -> 0

digitLen::Integer-> Int
digitLen = (length . map digitToInt . show)

isP::Int->Bool
isP n 
    |n < 2 = False
    |n == 2 = True
    |n `mod` 2 == 0 = False
    |otherwise = try n 3
    where
        try :: Int->Int->Bool
        try m i
            |i*i > m=True
            |m `mod` i == 0 = False
            |otherwise = try m (i+2)