CF1348A.Phoenix and Balance

Phoenix has $n$ coins with weights $2^1, 2^2, \dots, 2^n$. He knows that $n$ is even.

He wants to split the coins into two piles such that each pile has exactly $\frac{n}{2}$ coins and the difference of weights between the two piles is minimized. Formally, let $a$ denote the sum of weights in the first pile, and $b$ denote the sum of weights in the second pile. Help Phoenix minimize $|a-b|$, the absolute value of $a-b$.

凤凰有 $n$ 枚硬币，其重量分别为 $2^1, 2^2, \dots, 2^n$。他已知 $n$ 是偶数。

他希望将这些硬币分成两堆，使得每堆恰好包含 $\frac{n}{2}$ 枚硬币，并使两堆的总重量之差最小。形式化地说，设 $a$ 表示第一堆的总重量，$b$ 表示第二堆的总重量。请帮助凤凰最小化 $|a-b|$（即 $a-b$ 的绝对值）。

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains an integer $n$ ($2 \le n \le 30$; $n$ is even) — the number of coins that Phoenix has.

输入包含多个测试用例。第一行包含一个整数 $t$（$1 \le t \le 100$），表示测试用例的数量。

每个测试用例的第一行包含一个整数 $n$（$2 \le n \le 30$；且 $n$ 为偶数），表示凤凰拥有的硬币数量。

For each test case, output one integer — the minimum possible difference of weights between the two piles.

对于每个测试用例，输出一个整数——两堆重量之间的最小可能差值。

In the first test case, Phoenix has two coins with weights $2$ and $4$. No matter how he divides the coins, the difference will be $4-2=2$.

In the second test case, Phoenix has four coins of weight $2$, $4$, $8$, and $16$. It is optimal for Phoenix to place coins with weights $2$ and $16$ in one pile, and coins with weights $4$ and $8$ in another pile. The difference is $(2+16)-(4+8)=6$.

在第一个测试用例中，凤凰有两枚重量分别为 $2$ 和 $4$ 的硬币。无论他如何划分这些硬币，差值均为 $4-2=2$。

在第二个测试用例中，凤凰有四枚重量分别为 $2$、$4$、$8$ 和 $16$ 的硬币。最优策略是将重量为 $2$ 和 $16$ 的硬币放在一堆，将重量为 $4$ 和 $8$ 的硬币放在另一堆。此时差值为 $(2+16)-(4+8)=6$。