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) {
        Scanner sc = new Scanner(System.in);
        long N = sc.nextLong();
        sc.close();

        long rn = (long) Math.sqrt(N);
        ArrayList<Integer> ans = new ArrayList<>();
        for (long i = 1; i <= rn; i++) {
            if (N % i != 0) {
                continue;
            } else {
                long j = N / i;
                ans.add(dig(i, j));
            }
        }
        Collections.sort(ans);
        System.out.println(ans.get(0));
    }

    private static int dig(long a, long b) {
        int ad = 0;
        while (a > 0) {
            a /= 10;
            ad++;
        }
        int bd = 0;
        while (b > 0) {
            b /= 10;
            bd++;
        }
        return Math.max(ad, bd);
    }
}