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: Accepted)

object Main {
  def main(args: Array[String]): Unit = {
    import scala.io.StdIn.readLine
    val Array(datN) = readLine.split(" ").map(_.toLong)

    def makeDivisor(n: Long): Vector[Long] = {
      //for(i <- 1 to (math.sqrt(n).toInt + 1) if(i % 2 == 0)) yield i
      ((1 to (math.sqrt(n).toInt)).filter(i => n % i == 0)).map(_.toLong).toVector
    }
    val divN = makeDivisor(datN)

    def countDigit(a: Long): Int = {
      a.toString.split("").filterNot(_ == "").length
    }
    val ans = divN.foldLeft(100000)((acc, dat) =>
      math.min(acc, math.max(countDigit(dat), countDigit(datN / dat)))
    )

    println(ans)
  }
}