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 = {
    val sc = new java.util.Scanner(System.in)

    val N = sc.nextLong()

    // 約数列挙するのに O(√N) で間に合う
    // 列挙した約数の個数 M <= √N なので
    // 約数ごとに F(A, B) を 調べ その最小値を求める
    // O(M + √N) = O(√N). 間に合う

    def solve(N: Long): Int = {
      val upper = Math.sqrt(N)

      def F(a: Long, b: Long): Int =
        a.toString.length max b.toString.length

      def loop(a: Long, min: Int): Int =
        if (a <= upper) {
          if (N % a == 0) loop(a + 1, min min F(a, N / a))
          else loop(a + 1, min)
        } else min

      loop(1, 10)
    }

    println(solve(N))
  }

}