You are given a sequence of positive integers $A=(a_1,\ldots,a_N)$ of length $N$.  
There are $(2^N-1)$ ways to choose one or more terms of $A$. How many of them have an integer-valued average? Find the count modulo $998244353$.

Input is given from Standard Input in the following format:

```
$N$
$a_1$ $\ldots$ $a_N$
```

Print the answer.

-   $1 \leq N \leq 100$
-   $1 \leq a_i \leq 10^9$
-   All values in input are integers.

6

For each way to choose terms of A, the average is obtained as follows:

• If just $a_1$ is chosen, the average is $\frac{a_1}{1}=\frac{2}{1}=2$, which is an integer.

• If just $a_2$ is chosen, the average is $\frac{a_2}{1}=\frac{6}{1}=6$, which is an integer.

• If just $a_3$ is chosen, the average is $\frac{a_3}{1}=\frac{2}{1}=2$, which is an integer.

• If $a_1$ and $a_2$ are chosen, the average is $\frac{a_1+a_2}{2}=\frac{2+6}{2}=4$, which is an integer.

• If $a_1$ and $a_3$ are chosen, the average is $\frac{a_1+a_3}{2}=\frac{2+2}{2}=2$, which is an integer.

• If $a_2$ and $a_3$ are chosen, the average is $\frac{a_2+a_3}{2}=\frac{6+2}{2}=4$, which is an integer.

• If $a_1, a_2$, and $a_3$ are chosen, the average is $\frac{a_1+a_2+a_3}{3}=\frac{2+6+2}{3}=\frac{10}{3}$, which is not an integer.

Therefore, 6 ways satisfy the condition.

31

Regardless of the choice of one or more terms of A, the average equals 5.