CF1064B.Equations of Mathematical Magic

Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.

Arkadi and Boris Strugatsky. Monday starts on Saturday

Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: $a - (a \oplus x) - x = 0$ for some given $a$, where $\oplus$ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some $x$, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.

太棒了！——鹰钩鼻惊呼道。——一名程序员！这正是我们一直在寻找的人。

阿尔卡季·斯特鲁加茨基与鲍里斯·斯特鲁加茨基，《星期六开始的星期一》

罗曼·奥伊拉-奥伊拉和克里斯托瓦尔·洪塔在阅读《数学魔法方程》一书时，发现了一个有趣的方程：$a - (a \oplus x) - x = 0$，其中 $a$ 是给定的整数，$\oplus$ 表示两个整数的按位异或（XOR）运算（该运算在许多现代编程语言中记为 ^ 或 xor）。奥伊拉-奥伊拉迅速找到了某个满足该方程的解 $x$，但克里斯托瓦尔·洪塔认为奥伊拉-奥伊拉的结果还不够有趣，于是他问同事：这个方程有多少个非负整数解？这项任务对奥伊拉-奥伊拉而言过于困难，因此他请你来帮忙。

Each test contains several possible values of $a$ and your task is to find the number of equation's solution for each of them. The first line contains an integer $t$ ($1 \le t \le 1000$) — the number of these values.

The following $t$ lines contain the values of parameter $a$, each value is an integer from $0$ to $2^{30} - 1$ inclusive.

每个测试包含若干个可能的 $a$ 值，你的任务是针对每个 $a$ 值求出该方程的解的个数。第一行包含一个整数 $t$（$1 \le t \le 1000$），表示此类 $a$ 值的个数。

接下来的 $t$ 行每行包含一个参数 $a$ 的值，每个值均为介于 $0$ 到 $2^{30} - 1$（含端点）之间的整数。

For each value of $a$ print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of $a$ appear in the input.

One can show that the number of solutions is always finite.

对于每个 $a$ 的值，恰好输出一个整数——即给定参数值下该方程的非负解的个数。请按照输入中 $a$ 值出现的顺序输出答案。

可以证明，解的个数总是有限的。

Let's define the bitwise exclusive OR (XOR) operation. Given two integers $x$ and $y$, consider their binary representations (possibly with leading zeroes): $x_k \dots x_2 x_1 x_0$ and $y_k \dots y_2 y_1 y_0$. Here, $x_i$ is the $i$-th bit of the number $x$ and $y_i$ is the $i$-th bit of the number $y$. Let $r = x \oplus y$ be the result of the XOR operation of $x$ and $y$. Then $r$ is defined as $r_k \dots r_2 r_1 r_0$ where:

r\_i = \\left\\{ \\begin{aligned} 1, ~ \\text{if} ~ x\_i \\ne y\_i \\\\ 0, ~ \\text{if} ~ x\_i = y\_i \\end{aligned} \\right.

For the first value of the parameter, only $x = 0$ is a solution of the equation.

For the second value of the parameter, solutions are $x = 0$ and $x = 2$.

我们来定义按位异或（XOR）运算。给定两个整数 $x$ 和 $y$，考虑它们的二进制表示（允许有前导零）：$x_k \dots x_2 x_1 x_0$ 和 $y_k \dots y_2 y_1 y_0$。其中，$x_i$ 表示数 $x$ 的第 $i$ 位，$y_i$ 表示数 $y$ 的第 $i$ 位。令 $r = x \oplus y$ 为 $x$ 与 $y$ 的异或运算结果，则 $r$ 定义为 $r_k \dots r_2 r_1 r_0$，其中：

r\_i = \\left\\{ \\begin{aligned} 1, ~ \\text{若} ~ x\_i \\ne y\_i \\\\ 0, ~ \\text{若} ~ x\_i = y\_i \\end{aligned} \\right.

对于参数的第一个取值，方程的唯一解是 $x = 0$。

对于参数的第二个取值，方程的解为 $x = 0$ 和 $x = 2$。