Problem C
New Maths
“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 $\oplus $, is the same as normal addition, except any carries are ignored (in base $10$). Thus, $37 \oplus 48$ is $75$, not $85$.
Carryless multiplication, denoted by $\otimes $, is performed using the schoolboy algorithm for multiplication, column by column, but the intermediate additions are calculated using carryless addition. More formally, Let $a_ m a_{m-1} \ldots a_1 a_0$ be the digits of $a$, where $a_0$ is its least significant digit. Similarly define $b_ n b_{n-1} \ldots b_1 b_0$ be the digits of $b$. The digits of $c = a \otimes b$ are given by the following equation:
\[ c_ k = \left( a_0 b_ k \oplus a_1 b_{k-1} \oplus \cdots \oplus a_{k-1} b_1 \oplus a_ k b_0 \right) \bmod {10}, \]where any $a_ i$ or $b_ j$ is considered zero if $i > m$ or $j > n$. For example, $9 \otimes 1\, 234$ is $9\, 876$, $90 \otimes 1\, 234$ is $98\, 760$, and $99 \otimes 1\, 234$ is $97\, 536$.
Given $N$, find the smallest positive integer $a$ such that $a \otimes a = N$.
Input
The input consists of a single line with a positive integer $N$, with at most $25$ digits and no leading zeros.
Output
Print, on a single line, the least positive number $a$ such that $a \otimes a = N$. If there is no such $a$, print ‘-1’ instead.
| Sample Input 1 | Sample Output 1 |
|---|---|
6 |
4 |
| Sample Input 2 | Sample Output 2 |
|---|---|
149 |
17 |
| Sample Input 3 | Sample Output 3 |
|---|---|
123476544 |
11112 |
| Sample Input 4 | Sample Output 4 |
|---|---|
15 |
-1 |
