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 Ruby (2.3.3) (judge verdict: Accepted)

# 整数 N が与えられます。ここで、2 つの正の整数 A,B に対して、
# F(A,B) を「10 進表記における、A の桁数と B の桁数のうち大きい方」と定義します。
# 例えば、F(3,11) の値は、3 は 1 桁、11 は 2 桁であるため、F(3,11)=2 となります。
# 2 つの正の整数の組 (A,B) が N=A×B を満たすように動くとき、F(A,B) の最小値を求めてください。

n = gets.to_i

def comparison(i, j)
    return  i.to_s.length >= j.to_s.length ? i.to_s.length : j.to_s.length
end

if Math.sqrt(n) % 1 == 0
    puts Math.sqrt(n).to_i.to_s.length
else
    num = Math.sqrt(n).to_i
    while n % num != 0
        num -= 1
    end
    puts (n / num).to_s.length >= num.to_s.length ? (n / num).to_s.length : num.to_s.length
end