This is a non-standard but still positional **numeral system**. Its feature is that digits can have one of the values `-1`

, `0`

and `1`

.
Nevertheless, its base is still `3`

(because there are three possible values). Since it is not convenient to write `-1`

as a digit,
we'll use letter `Z`

further for this purpose. If you think it is quite a strange system - look at the picture - here is one of the
computers utilizing it.

So here are few first numbers written in balanced ternary:

```
0 0
1 1
2 1Z
3 10
4 11
5 1ZZ
6 1Z0
7 1Z1
8 10Z
9 100
```

This system allows you to write negative values without leading minus sign: you can simply invert digits in any positive number.

```
-1 Z
-2 Z1
-3 Z0
-4 ZZ
-5 Z11
```

Note that a negative number starts with `Z`

and positive with `1`

.

It is easy to represent a given number in **balanced ternary** via temporary representing it in normal ternary number system. When value is
in standard ternary, its digits are either `0`

or `1`

or `2`

. Iterating from the lowest digit we can safely skip any `0`

s and `1`

s,
however `2`

should be turned into `Z`

with adding `1`

to the next digit. Digits `3`

should be turned into `0`

on the same terms -
such digits are not present in the number initially but they can be encountered after increasing some `2`

s.

**Example 1:** Let us convert `64`

to balanced ternary. At first we use normal ternary to rewrite the number:

$$ 64_{10} = 02101_{3} $$

Let us process it from the least significant (rightmost) digit:

`1`

,`0`

and`1`

are skipped as it is.( Because`0`

and`1`

are allowed in balanced ternary )`2`

is turned into`Z`

increasing the digit to its left, so we get`1Z101`

.

The final result is `1Z101`

.

Let us convert it back to the decimal system by adding the weighted positional values: $$ 1Z101 = 81 \cdot 1 + 27 \cdot (-1) + 9 \cdot 1 + 3 \cdot 0 + 1 \cdot 1 = 64_{10} $$

**Example 2:** Let us convert `237`

to balanced ternary. At first we use normal ternary to rewrite the number:

$$ 237_{10} = 22210_{3} $$

Let us process it from the least significant (rightmost) digit:

`0`

and`1`

are skipped as it is.( Because`0`

and`1`

are allowed in balanced ternary )`2`

is turned into`Z`

increasing the digit to its left, so we get`23Z10`

.`3`

is turned into`0`

increasing the digit to its left, so we get`30Z10`

.`3`

is turned into`0`

increasing the digit to its left( which is by default`0`

), and so we get`100Z10`

.

The final result is `100Z10`

.

Let us convert it back to the decimal system by adding the weighted positional values: $$ 100Z10 = 243 \cdot 1 + 81 \cdot 0 + 27 \cdot 0 + 9 \cdot (-1) + 3 \cdot 1 + 1 \cdot 0 = 237_{10} $$