We are given a small Python program which computes the flag:

```
def crunchy(n):
if n < 2: return n
return 6 * crunchy(n - 1) + crunchy(n - 2)
= ... # a _really_ big number
g print("Your flag is: INSA{%d}"%(crunchy(g)%100000007))
```

Running this, of course, wonβt get us anywhere with the exponentially slow implementation of `crunchy`

. A simple improvement would be to cache computed vaues of `crunchy(n)`

, but that reduces the complexity to linear, which is still not nearly enough with the hundred-digit parameter.

Since `crunchy`

is a linear recurrence, we can compute it in logarithmic time. We can express the mapping from $(c(n), c(n-1))$ to $(c(n+1), c(n))$ (where $c$ is the function) as a matrix multiplication:

$\begin{pmatrix}c(n+1)\\c(n)\end{pmatrix} = \begin{pmatrix}6&1\\1&0\end{pmatrix} \begin{pmatrix}c(n)\\c(n-1)\end{pmatrix}$

If we expand the right-hand side of this equation $n-1$ times, we will get:

$\begin{pmatrix}c(n+1)\\c(n)\end{pmatrix} = \begin{pmatrix}6&1\\1&0\end{pmatrix}^n \begin{pmatrix}c(1)\\c(0)\end{pmatrix}$

Since we know the values of $c(1)$ and $c(0)$, all that remains is to compute the exponentiated matrix. We can do this in $\mathcal{O}(\log{n})$ multiplications by repeated squaring, which is completely reasonable for the input value of `g`

(a few hundreds of multiplications).

Note that all multiplications should be done modulo $10^9 + 7$ instead of only doing so at the end β otherwise, the intermediate values will be to large to work with (they grow exponentially).