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Codeforces
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Codeforces Round 792 (Div. 1 + Div. 2)
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The problem statement has recently been changed.View the changes.
×
B. Z mod X = C
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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

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

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.

Input

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.

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

Output

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

You can output any correct answer.

Example
Input
41 3 4127 234 4212 7 859 94 388
Output
12 11 41063 234 148425 23 82221 94 2609
Note

In the first test case:

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

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

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


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