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 dd = divisor(n)

    var min = Long.MAX_VALUE

    dd.forEach {
        val d1 = it
        val d2 = n / d1

        val l1 = d1.toString().length.toLong()
        val l2 = d2.toString().length.toLong()

        min = Math.min(min, Math.max(l1, l2))
    }

    println(min)
}

fun divisor(value : Long) : List<Long> {
    val divisors = mutableListOf<Long>()
    var i = 1L
    while (i * i <= value) {
        if (0L == value % i) {
            divisors.add(i)
            if (i != value/i) divisors.add(value/i)
        }
        i++
    }
    return divisors.sorted()
}