A **bitstring** is a string consisting of 0s and 1s. However, you’re only looking for bitstrings with the following properties:

- There are no two consecutive 1s in the bitstring.
- Every run of 0s is of even length (i.e. every block of 0s has an even number of 0s in it).

1001 is an example of such a bitstring, but 10001 is not. Luckily, your computer science (or combinatorics) teacher shares a formula for figuring out how many such bitstrings exist for any given length `n`:

`s`(0) = 1,`s`(1) = 1,`s`(2) = 1.`s`(`n`) =`s`(`n`−2) + s(`n`−3) for all`n`> 2.

That is, there is only 1 string of size 0 (empty string matches both rules), only 1 string of size 1 (“1”), and only 1 string of size 2 (“00”). For size 3, you’d need to calculate the sum of `s`(3−2) and `s`(3−3), which are known from the results above.

The input file **DATA3.txt** will contain 5 test cases, each being a line with a single integer 1 ≤ `n` ≤ 75, the length of the bitstring.

The output file **OUT3.txt** will contain 5 lines of output, each being the number of different bitstrings of the corresponding length `n` with the described properties.

```
1
20
```

```
1
200
```