CF1379B.Dubious Cyrpto

Pasha loves to send strictly positive integers to his friends. Pasha cares about security, therefore when he wants to send an integer $n$, he encrypts it in the following way: he picks three integers $a$, $b$ and $c$ such that $l \leq a,b,c \leq r$, and then he computes the encrypted value $m = n \cdot a + b - c$.

Unfortunately, an adversary intercepted the values $l$, $r$ and $m$. Is it possible to recover the original values of $a$, $b$ and $c$ from this information? More formally, you are asked to find any values of $a$, $b$ and $c$ such that

• $a$, $b$ and $c$ are integers,
• $l \leq a, b, c \leq r$,
• there exists a strictly positive integer $n$, such that $n \cdot a + b - c = m$.

帕沙喜欢向朋友发送严格正整数。帕沙注重安全性，因此当他想要发送一个整数 $n$ 时，他会以如下方式对其进行加密：他选取三个整数 $a$、$b$ 和 $c$，满足 $l \leq a,b,c \leq r$，然后计算加密后的值 $m = n \cdot a + b - c$。

不幸的是，一名攻击者截获了 $l$、$r$ 和 $m$ 这三个值。能否仅凭这些信息恢复出原始的 $a$、$b$ 和 $c$？更准确地说，你需要找出任意一组满足以下条件的 $a$、$b$ 和 $c$：

• $a$、$b$ 和 $c$ 均为整数，
• $l \leq a, b, c \leq r$，
• 存在某个严格正整数 $n$，使得 $n \cdot a + b - c = m$。

The first line contains the only integer $t$ ($1 \leq t \leq 20$) — the number of test cases. The following $t$ lines describe one test case each.

Each test case consists of three integers $l$, $r$ and $m$ ($1 \leq l \leq r \leq 500\,000$, $1 \leq m \leq 10^{10}$). The numbers are such that the answer to the problem exists.

第一行包含一个整数 $t$（$1 \leq t \leq 20$），表示测试用例的数量。接下来的 $t$ 行每行描述一个测试用例。

每个测试用例包含三个整数 $l$、$r$ 和 $m$（$1 \leq l \leq r \leq 500\,000$，$1 \leq m \leq 10^{10}$）。所给数据保证问题有解。

For each test case output three integers $a$, $b$ and $c$ such that, $l \leq a, b, c \leq r$ and there exists a strictly positive integer $n$ such that $n \cdot a + b - c = m$. It is guaranteed that there is at least one possible solution, and you can output any possible combination if there are multiple solutions.

对于每个测试用例，输出三个整数 $a$、$b$ 和 $c$，满足 $l \leq a, b, c \leq r$，且存在一个严格正整数 $n$，使得 $n \cdot a + b - c = m$。题目保证至少存在一个可行解；若存在多个解，可输出任意一组可行解。

In the first example $n = 3$ is possible, then $n \cdot 4 + 6 - 5 = 13 = m$. Other possible solutions include: $a = 4$, $b = 5$, $c = 4$ (when $n = 3$); $a = 5$, $b = 4$, $c = 6$ (when $n = 3$); $a = 6$, $b = 6$, $c = 5$ (when $n = 2$); $a = 6$, $b = 5$, $c = 4$ (when $n = 2$).

In the second example the only possible case is $n = 1$: in this case $n \cdot 2 + 2 - 3 = 1 = m$. Note that, $n = 0$ is not possible, since in that case $n$ is not a strictly positive integer.

在第一个例子中，$n = 3$ 是可行的，此时 $n \cdot 4 + 6 - 5 = 13 = m$。其他可行解包括：$a = 4$，$b = 5$，$c = 4$（当 $n = 3$ 时）；$a = 5$，$b = 4$，$c = 6$（当 $n = 3$ 时）；$a = 6$，$b = 6$，$c = 5$（当 $n = 2$ 时）；$a = 6$，$b = 5$，$c = 4$（当 $n = 2$ 时）。

在第二个例子中，唯一可行的情况是 $n = 1$：此时 $n \cdot 2 + 2 - 3 = 1 = m$。注意，$n = 0$ 不可行，因为此时 $n$ 不是严格正整数。