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 void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        
        long a = 1;
        
        for(long i = 1; i * i <= n; i++) {
            if(n % i == 0) {
                if(Math.abs(i - n/i) < Math.abs(a - n/a))
                    a = i;
            }
        }
        int cntA = 0;
        long A = a;
        while(A > 0) {
            cntA++;
            A /= 10;
        }
        
        int cntN = 0;
        long N = n/a;
        while(N > 0) {
            cntN++;
            N /= 10;
        }
		System.out.println(Math.max(cntN, cntA));
    }
    
}