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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
  val sc = Scanner(System.`in`)
  val n = sc.nextLong()

  println(abc057c(n))
}

private fun abc057c(n: Long): Int {
  val sqrt = Math.sqrt(n.toDouble()).toInt()

  val pairList = mutableListOf<Pair<Long, Long>>()
  for (i in 1..sqrt) {
    if (n % i == 0L) {
      val pair = Pair(i.toLong(), n / i)
      pairList.add(pair)
    }
  }

  val digitList = mutableListOf<Int>()
  for (pi in pairList) {
    val f = pi.first
    val s = pi.second
    val d = Math.max(f.toString().length, s.toString().length)
    digitList.add(d)
  }
  val ans = digitList.min()!!

  return ans
}