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: Wrong Answer)

use std::str::FromStr;

fn divisor(n : u64) -> Vec<u64> {
    let mut x = n;
    let mut v = Vec::new();
    let mut i : u64 = 2;
    while i.pow(2) <= n {
        while x % i == 0 {
            v.push(i);
            x = x / i;
        }
        i += 1;
    }
    if x != 1 {
        v.push(x);
    }
    return v;
}

fn get_one<T : FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().parse().ok().unwrap();
}

fn get_vec<T : FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();
}

fn main() {
    let n : u64 = get_one();
    let mut v = divisor(n);
    v.sort();
    v.reverse();
    //println!("{:?}", v);
    let mut a = 1;
    let mut b = 1;
    for x in v {
        if a < b {
            a *= x;
        } else {
            b *= x;
        }
    }
    //println!("{} {}", a, b);
    let mut y = std::cmp::max(a,b);
    let mut ans = 0;
    while y > 0 {
        ans += 1;
        y = y / 10;
    }

    println!("{}", ans);
}