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: Wrong Answer)

fun main(args: Array<String>) {
    var numN =  readLine().orEmpty().toLong()
    val sqrtN = Math.sqrt(numN.toDouble()).toLong()

    val listA = mutableListOf<Long>()
    
    for(i in 1L until sqrtN){
        if(numN % i == 0L){
            listA.add(i)
        }
    }

    //println(partN.distinct())
    var dist = numN
    var result = 1L
    listA.distinct().forEach {
        if(Math.abs(it - (numN / it)) < dist){
            dist = Math.abs(it - (numN / it))
            result = it
        }
    }

    println(digit(result,numN / result))

}


//桁数の大きい方の値
fun digit(a:Long,b:Long):Int{
    val digitA = a.toString().length
    val digitB = b.toString().length

    return if (digitA < digitB) digitB else digitA
}