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.io.*;
import java.math.*;
import java.util.*;
 
public class Main {
    private static boolean debug = false;
    private static boolean elapsed = false;
 
    private static PrintWriter _out = new PrintWriter(System.out);
    private static PrintWriter _err = new PrintWriter(System.err);
 
    private void solve(Scanner sc) {
        long N = sc.nextLong();
 
        int ans = Integer.MAX_VALUE;
        for (long A = 1; A <= Math.sqrt(N); ++A) {
            if (N % A != 0) {
                continue;
            }
            long B = N / A;
            ans = Math.min(ans, F(A, B));
        }
        _out.println(ans);
    }
    private int F(long a, long b) {
        return Math.max(String.valueOf(a).length(), String.valueOf(b).length());
    }
    private static BigInteger C(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        for (long i = r; i > 1; --i) {
            res = res.divide(BigInteger.valueOf(i));
        }
        return res;
    }
    private static BigInteger P(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        return res;
    }
    /*
     * 10^10 > Integer.MAX_VALUE = 2147483647 > 10^9
     * 10^19 > Long.MAX_VALUE = 9223372036854775807L > 10^18
     */
    public static void main(String[] args) {
        long S = System.currentTimeMillis();
 
        Scanner sc = new Scanner(System.in);
        new Main().solve(sc);
        _out.flush();
 
        long G = System.currentTimeMillis();
        if (elapsed) {
            _err.println((G - S) + "ms");
        }
        _err.flush();
    }
}