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: Accepted)

import Data.List

main = interact $ show . sol . read

sol n | n<=9 = 1
sol n = succ . floor $ log (fromIntegral (m+1)) / log 10
  where
  q = ceiling . sqrt $ fromIntegral n
  ps = fmap ((,) <$> head <*> length) . group $ primeFactors n
  ds = foldr1 ((<*>) . ((*) <$>)) $ fmap (\x -> (fst x^) <$> [0..snd x]) ps
  m = minimum [d | d <- ds, d>=q]

-- | Prime numbers
primes :: Integral a => [a]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]

primeFactors :: Integral a => a -> [a]
primeFactors 1 = []
primeFactors n = factor n primes

factor :: Integral a => a -> [a] -> [a]
factor n (p:ps) 
  | p*p > n        = [n]
  | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
  | otherwise      = factor n ps