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 Rust (1.15.1) (judge verdict: Accepted)

use std::cmp::max;

fn get_line() -> String {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).unwrap();
    line.trim().to_string()
}

fn digit(n: u64) -> u64 {
    if n < 10 {
        return 1;
    }
    digit(n / 10) + 1
}

fn max_digit(a: u64, b: u64) -> u64 {
    max(digit(a), digit(b))
}

fn main() {
    let n = get_line().parse::<u64>().unwrap();
    let sqrt_n = (n as f32).sqrt() as u64;
    let mut result = 100000;
    for a in 1..sqrt_n + 1 {
        if n % a != 0 { continue; }
        let b = n / a;
        let f = max_digit(a, b);
        if f < result {
            result = f;
        }
    }
    println!("{}", result);
}