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)

prime_factors :: Int -> [Int]

prime_factors 1 = []
prime_factors n
  | factors == []  = [n]
  | otherwise = factors ++ prime_factors (n `div` (head factors))
  where factors = take 1 $ filter (\x -> (n `mod` x) == 0) [2 .. n-1]

divide::[Int]->(Int,Int)->(Int,Int)
divide [] (a,b)=(a,b)
divide (x:xs) (a,b) = if a>=b then divide xs (a,b*x) else divide xs (a*x,b)

find_ab::Int->(Int,Int)
find_ab  n = let fac = reverse (prime_factors n) in
             divide fac (1,1)

dig::Int -> Int
dig x = (truncate (logBase 10.0 (fromIntegral x)))+1

solver::Int -> Int
solver n = let (a,b) =  find_ab n in max (dig a) (dig b)

main::IO()
main=do
  nc<-getLine
  print (solver (read nc))