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 (1.42.0) (judge verdict: Accepted)

fn main() {
    proconio::input! {
        n: usize,
    }
    fn factor(mut n: usize) -> Vec<(usize, usize)> {
        let mut ret = vec![];
        for i in 2.. {
            if i * i > n {
                break;
            }
            let mut cnt = 0;
            while n % i == 0 {
                cnt += 1;
                n /= i;
            }
            if cnt != 0 {
                ret.push((i, cnt));
            }
        }
        if n != 1 {
            ret.push((n, 1));
        }
        ret
    }
    let factors = factor(n);
    fn search(factor_idx: usize, v: Vec<usize>, factors: &[(usize, usize)]) -> usize {
        fn pow(u: usize, i: usize) -> usize {
            (0..i).fold(1, |sum, _| sum * u)
        }
        fn number_of_digits(u: usize) -> usize {
            let mut u = u;
            let mut count = 1;
            while u >= 10 {
                u /= 10;
                count += 1;
            }
            count
        }
        if factor_idx >= factors.len() {
            let nx = (0..factors.len()).fold(1, |sum, i| {
                sum * pow(factors[i].0, *v.get(i).or(Some(&1)).unwrap())
            });
            let ny = (0..factors.len()).fold(1, |sum, i| {
                sum * pow(factors[i].0, factors[i].1 - *v.get(i).or(Some(&1)).unwrap())
            });
            let x = number_of_digits(nx);
            let y = number_of_digits(ny);
            std::cmp::max(x, y)
        } else {
            let mut min = std::usize::MAX;
            for i in factor_idx..factors.len() {
                for j in 0..factors[i].1 {
                    let mut v = v.clone();
                    v.push(j);
                    min = std::cmp::min(min, search(i + 1, v, factors));
                }
            }
            min
        }
    }
    let ans = search(0, vec![], &factors);
    println!("{}", ans);
}