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: Runtime Error)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {
		Scanner scanner = new Scanner(System.in);
		long n=scanner.nextLong();
		List<Long> list = new ArrayList<Long>();
		list.add((long)1);
		while(n % 2 == 0){
			n = n / 2;
			list.add((long) 2);
		}

		for(int i = 3; i*i-1 <= n; i += 2){
			while(n % i == 0){
				n = n / i;
				list.add((long) i);
			}
		}
		if(n > 1){
			list.add(n);
		}

		list.sort(Comparator.reverseOrder());

		long l = list.get(0);
		long s = list.get(1);

		for(int i = 2; i < list.size(); i++){
			s = s * list.get(i);
			if(l < s){
				long temp = l;
				l = s;
				s = temp;
			}
		}

		int p = String.valueOf(l).length();
		int q = String.valueOf(s).length();

		if(p < q){
			System.out.println(q);
		}else{
			System.out.println(p);
		}
	}
}