Given is a positive integer $N$, where none of the digits is $0$.  
Let $k$ be the number of digits in $N$. We want to make a multiple of $3$ by erasing at least $0$ and at most $k-1$ digits from $N$ and concatenating the remaining digits without changing the order.  
Determine whether it is possible to make a multiple of $3$ in this way. If it is possible, find the minimum number of digits that must be erased to make such a number.

Input is given from Standard Input in the following format:

```
$N$
```

If it is impossible to make a multiple of $3$, print `-1`; otherwise, print the minimum number of digits that must be erased to make such a number.

-   $1 \le N \lt 10^{18}$
-   None of the digits in $N$ is $0$.

1

By erasing the 5, we get the number 3, which is a multiple of 3. Here we erased the minimum possible number of digits - 1.

0

Note that we can choose to erase no digit.

1

For example, by erasing the 8, we get the number 622734, which is a multiple of 3.

-1

Note that we must erase at least 0 and at most k−1 digits, where k is the number of digits in N, so we cannot erase all the digits.
In this case, it is impossible to make a multiple of 3 in the way described in the problem statement, so we should print -1.