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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
	    Scanner scanner = new Scanner(System.in);

	    long N = Long.parseLong(scanner.next());
	    long max = (long)N;
	    int c = Integer.MAX_VALUE;
	    for (long i = 1; i <= max; i++) {
	        if ((N % i) == 0) {
	            long a = i;
	            long b = (long)N / i;
	            if (a > b) break;
	            int ad = getDigit(a);
	            int bd = getDigit(b);
	            int md = Math.max(ad, bd);
	            max = Math.max(a, b);
	            if (md < c) {
	                c = md;
	            }
	        } else {
	            max = N / i + 1;
	        }
	    }

        System.out.print(c);
	}

	public static int getDigit(long n) {
	    int c = 0;
		while (n >= 1)
	    {
			c++;
	        n = n/10;
	    }
		return c;
	}
}