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.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Main {
	
	public static void main(String[] args) throws IOException {
		final Scanner sc = new Scanner(System.in);
		
		final long N = sc.nextLong();
		
		int min = Integer.MAX_VALUE;
		for(long j = 1; j * j <= N; j++){
			if(N % j != 0){ continue; }
			
			min = Math.min(min, Math.max(Long.toString(j).length(), Long.toString(N / j).length()));
		}
		
		System.out.println(min);
	}
	
	public static class Scanner implements Closeable {
		private BufferedReader br;
		private StringTokenizer tok;
 
		public Scanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
		}
 
		private void getLine() throws IOException {
			while (!hasNext()) {
				tok = new StringTokenizer(br.readLine());
			}
		}
 
		private boolean hasNext() {
			return tok != null && tok.hasMoreTokens();
		}
 
		public String next() throws IOException {
			getLine();
			return tok.nextToken();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
		
		public int[] nextIntArray(int n) throws IOException {
			final int[] ret = new int[n];
			for(int i = 0; i < n; i++){
				ret[i] = this.nextInt();
			}
			return ret;
		}
		
		public long[] nextLongArray(int n) throws IOException {
			final long[] ret = new long[n];
			for(int i = 0; i < n; i++){
				ret[i] = this.nextLong();
			}
			return ret;
		}
 
		public void close() throws IOException {
			br.close();
		}
	}
}