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)

fun main(args: Array<String>) {
    val N = readLine()!!.toLong()


    val sqrtN = Math.sqrt(N.toDouble()).toLong()

    // log_10(x)
    fun digits(x: Long) : Int {
        var num = x
        var count = 0
        while (num > 0) {
            num /= 10
            count++
        }
        return count
    }

    fun F(A: Long, B: Long) : Int {
        return Math.max(digits(A), digits(B))
    }

    var answer = Int.MAX_VALUE
    // O(N^0.5 * log_10 N)
    (1..sqrtN).forEach { A ->
        if (N % A == 0L) {
            val B = N / A
            answer = Math.min(answer, F(A, B))
        }
    }
    println(answer)
}