CF1551A.Polycarp and Coins

Polycarp must pay exactly $n$ burles at the checkout. He has coins of two nominal values: $1$ burle and $2$ burles. Polycarp likes both kinds of coins equally. So he doesn't want to pay with more coins of one type than with the other.

Thus, Polycarp wants to minimize the difference between the count of coins of $1$ burle and $2$ burles being used. Help him by determining two non-negative integer values $c_1$ and $c_2$ which are the number of coins of $1$ burle and $2$ burles, respectively, so that the total value of that number of coins is exactly $n$ (i. e. $c_1 + 2 \cdot c_2 = n$), and the absolute value of the difference between $c_1$ and $c_2$ is as little as possible (i. e. you must minimize $|c_1-c_2|$).

波利卡普在结账时必须恰好支付 $n$ 布尔。他有两种面额的硬币：$1$ 布尔和 $2$ 布尔。波利卡普对这两种硬币同样喜爱，因此他不希望使用其中一种硬币的数量多于另一种。

于是，波利卡普希望最小化所用的 $1$ 布尔硬币数量与 $2$ 布尔硬币数量之间的差值。请帮助他确定两个非负整数 $c_1$ 和 $c_2$，分别表示所用的 $1$ 布尔硬币数和 $2$ 布尔硬币数，使得这些硬币的总价值恰好为 $n$（即满足 $c_1 + 2 \cdot c_2 = n$），且 $c_1$ 与 $c_2$ 的差的绝对值尽可能小（即最小化 $|c_1-c_2|$）。

The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases. Then $t$ test cases follow.

Each test case consists of one line. This line contains one integer $n$ ($1 \le n \le 10^9$) — the number of burles to be paid by Polycarp.

第一行包含一个整数 $t$（$1 \le t \le 10^4$）—— 测试用例的数量。接下来是 $t$ 个测试用例。

每个测试用例由一行组成。该行包含一个整数 $n$（$1 \le n \le 10^9$）—— Polycarp 需要支付的 burles 数量。

For each test case, output a separate line containing two integers $c_1$ and $c_2$ ($c_1, c_2 \ge 0$) separated by a space where $c_1$ is the number of coins of $1$ burle and $c_2$ is the number of coins of $2$ burles. If there are multiple optimal solutions, print any one.

对于每个测试用例，输出一行，包含两个整数 $c_1$ 和 $c_2$（$c_1, c_2 \ge 0$），以空格分隔，其中 $c_1$ 表示面值为 $1$ burle 的硬币数量，$c_2$ 表示面值为 $2$ burles 的硬币数量。若存在多个最优解，输出任意一个即可。

The answer for the first test case is "334 333". The sum of the nominal values of all coins is $334 \cdot 1 + 333 \cdot 2 = 1000$, whereas $|334 - 333| = 1$. One can't get the better value because if $|c_1 - c_2| = 0$, then $c_1 = c_2$ and $c_1 \cdot 1 + c_1 \cdot 2 = 1000$, but then the value of $c_1$ isn't an integer.

The answer for the second test case is "10 10". The sum of the nominal values is $10 \cdot 1 + 10 \cdot 2 = 30$ and $|10 - 10| = 0$, whereas there's no number having an absolute value less than $0$.

第一个测试用例的答案是“334 333”。所有硬币面值的总和为 $334 \cdot 1 + 333 \cdot 2 = 1000$，而 $|334 - 333| = 1$。无法得到更优的结果，因为若 $|c_1 - c_2| = 0$，则 $c_1 = c_2$，此时有 $c_1 \cdot 1 + c_1 \cdot 2 = 1000$，但该方程解出的 $c_1$ 不是整数。

第二个测试用例的答案是“10 10”。面值总和为 $10 \cdot 1 + 10 \cdot 2 = 30$，且 $|10 - 10| = 0$；而绝对值不可能小于 $0$。