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 Scala (2.11.7) (judge verdict: Wrong Answer)

import io.StdIn.readLong
import annotation.tailrec

object Main extends App {

  def isPrime(n: Long) = {
    val pfcsOf = (n: Long) => if (n < 4) Nil else 2L :: (3L to math.sqrt(n).toLong by 2).toList
    if (n < 2) false
    else pfcsOf(n).forall(n % _ != 0)
  }

  val primes = 2L #:: Stream.from(3, 2).map(_.toLong).filter(isPrime)

  @tailrec
  def primeFactorize(n: Long, pfs: List[Long]): List[Long] = {
    if (n == 1) pfs
    else {
      val Some(p) = primes.find(n % _ == 0)
      primeFactorize(n / p, p :: pfs)
    }
  }

  def digitOf(n: Long) = math.log10(n).floor.toInt + 1

  val n = readLong
  val dn = digitOf(n)
  val pfs = primeFactorize(n, Nil).sorted

  val cands = dn +: (for (i <- 1 to pfs.size) yield {
    digitOf(n / pfs.take(i).product)
  })

  println(cands.filter(_ >= (dn + 1) /2).min)
}