题号：NC224724
时间限制：C/C++/Rust/Pascal 1秒，其他语言2秒
空间限制：C/C++/Rust/Pascal 256 M，其他语言512 M
64bit IO Format: %lld
时间限制：C/C++/Rust/Pascal 1秒，其他语言2秒
空间限制：C/C++/Rust/Pascal 256 M，其他语言512 M
64bit IO Format: %lld
题目描述
Carryless addition is the same as normal addition, except any carries are ignored (in base 10). Thus, 37 + 48 is 75, not 85.
Carryless multiplication is performed using the schoolbook algorithm for multiplication, column by column, but the intermediate sums are calculated using carryless addition. Thus:
9 ∙ 1234 = 9000 + (900 + 900) + (90 + 90 + 90) + (9 + 9 + 9 + 9) = 9000 + 800 + 70 + 6 = 9876
90 ∙ 1234 = 98760
99 ∙ 1234 = 98760 + 9876 = 97536
Formally, define 
to be the 
digit of the value 𝑐. If 𝑐 = 𝑎 ∙ 𝑏 then 
Given an integer 𝒏, calculate the smallest positive integer 𝒂 such that 𝒂 ∙ 𝒂 = 𝒏 in carryless multiplication.
输入描述:
The input consists of a single line with an integer n ().
输出描述:
Output the smallest positive integer that is a carryless square root of the input number, or −1 if no such number exists.