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)

# 約数を列挙する。配列形式で返す。
def divisors(n)
    ret = []
    i = 1
    while i*i <= n do
        if n % i == 0
            ret.push i
            ret.push n/i unless i*i == n
        end
        i += 1
    end
    ret.sort
end

# nの桁数を数える。 n<=0のときは0を返す。
def digits_cnt(n)
    cnt = 0
    ni = n
    while ni > 0 do
        ni /= 10
        cnt += 1
    end
    return cnt
end

# p divisors(gets.to_i)
# p digits_cnt(gets.to_i)

N = gets.to_i
i = 1
min_fab = 9999
while i*i <= N do
    if N % i == 0
        a = i
        b = N/i
        fab = [digits_cnt(a), digits_cnt(b)].max
        min_fab = fab if min_fab > fab
    end
    i += 1
end
puts min_fab