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 (1.3.71) (judge verdict: Accepted)

import kotlin.math.sqrt

fun main() {
    var n = readLine()!!.toLong()
    if (n < 10) {
        print(1)
        return
    }

    val x = sqrt(n.toDouble()).toInt()

    var i = 1L

    var divs = mutableListOf<Int>()
    var answer = Int.MAX_VALUE
    while (i <= x) {
        if ( n % i == 0L) {
            val a= i.toString().toCharArray().size
            val b= (n/i).toString().toCharArray().size
            val max = if (a < b ) b else a

            if (max < answer) {
                answer = max
            }
        }
        i+= 1
    }

    print(answer)

}