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 = sc.nextLong();
        int r = (int) Math.ceil(Math.sqrt(n));
        long tmp = n;
        long a = 1;
        long b = 1;
        for (int i = 2; i < r; i++) {
            while(tmp % i == 0){
                tmp /= i;
                if(a > b) b *= i;
                else a *= i;
            }
        }
        int ans;
        if(tmp == n) ans = String.valueOf(n).length();
        else if(tmp >= r) ans = String.valueOf(tmp).length();
        else if(a > b) ans = String.valueOf(a).length();
        else ans = String.valueOf(b).length();

        System.out.println(ans);
        sc.close();
    }

}