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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
 
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long n = Long.parseLong(sc.next());
		long a = 1;
		long b = n;
		int min = String.valueOf(n).length();
		if(Math.sqrt(n) - (int)(Math.sqrt(n)) == 0) {
		    min = String.valueOf(Math.sqrt(n)).length();
		}else {
    		for(int i=2; i<=Math.sqrt(n); i++) {
    		    if(b % i == 0) {
    		        a *= i;
    		        b /= i;
    		        if(min > Math.max(String.valueOf(a).length(), String.valueOf(b).length())) {
    		            min = Math.max(String.valueOf(a).length(), String.valueOf(b).length());
    		        }
    		        i--;
    		    }
    		}
		}
		
		System.out.println(min);
	}
}