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
空间限制：C/C++/Rust/Pascal 256 M，其他语言512 M
64bit IO Format: %lld
题目描述
“Drat!” cursed Charles. “This stupid carry bar is not working in my Engine! I just tried to calculate the square of a number, but it’s wrong; all of the carries are lost.”
“Hmm,” mused Ada, “arithmetic without carries! I wonder if I can figure out what your original input was, based on the result I see on the Engine.”
Carryless addition, denoted by ⊕, is the same as normal addition, except any carries are ignored (in base 10). Thus, 37⊕48 is 75, not 85.
Carryless multiplication, denoted by ⊗, is performed using the schoolboy algorithm for multiplication, column by column, but the intermediate additions are calculated using carryless addition. More formally, Let 
be the digits of a, where 
is its least significant digit. Similarly define 
be the digits of b. The digits of c=a⊗b are given by the following equation:
where any or
is considered zero if i>m or j>n. For example, 9⊗1234 is 98769876, 90⊗1234 is 9876098760, and 99⊗1234 is 9753697536.
Given N, find the smallest positive integer aa such that a⊗a=N.
输入描述:
The input consists of a single line with a positive integer N, with at most 25 digits and no leading zeros.
输出描述:
Print, on a single line, the least positive number aa such that a⊗a=N. If there is no such a, print ‘-1’ instead.