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: Accepted)

import java.util.*;
public class Main{
	public static int order(long a){
		int o=0;
		long d=1;
		while(d<=a){
			o++;d*=10;
		}
		return o;
	}
	public static int maxOrder(long a, long b){
		return Math.max(order(a), order(b));
	}
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		long N =Long.parseLong(sc.next());
		int minF=order(N);
		for (long i=1;i*i<=N;i++){
			if(N%i==0){
				minF=Math.min(minF,maxOrder(i,N/i));
			}
		}
		System.out.println(minF);
	}
}