Codeforces Round 792 (Div. 1 + Div. 2)
B
Z mod X = C
Codeforces
800
原题
← AC →
翻译中…

You are given three positive integers $a$, $b$, $c$ ($a \lt b \lt c$). You have to find three positive integers $x$, $y$, $z$ such that:

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$$x \bmod y = a,$$ $$y \bmod z = b,$$ $$z \bmod x = c.$$

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Here $p \bmod q$ denotes the remainder from dividing $p$ by $q$. It is possible to show that for such constraints the answer always exists.

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Input

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The input consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10\,000$) — the number of test cases. Description of the test cases follows.

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Each test case contains a single line with three integers $a$, $b$, $c$ ($1 \le a \lt b \lt c \le 10^8$).

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Output

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For each test case output three positive integers $x$, $y$, $z$ ($1 \le x, y, z \le 10^{18}$) such that $x \bmod y = a$, $y \bmod z = b$, $z \bmod x = c$.

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You can output any correct answer.

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Note

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In the first test case:

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$$x \bmod y = 12 \bmod 11 = 1;$$

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$$y \bmod z = 11 \bmod 4 = 3;$$

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$$z \bmod x = 4 \bmod 12 = 4.$$

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样例

Input #1
4
1 3 4
127 234 421
2 7 8
59 94 388
Output #1
12 11 4
1063 234 1484
25 23 8
2221 94 2609
InputCopy OutputCopy
题目信息
题号
1684B
时间限制
1 s
内存限制
256 MB
标签
构造
数学

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