C. Trip to the Olympiad
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

In the upcoming year, there will be many team olympiads, so the teachers of "T-generation" need to assemble a team of three pupils to participate in them. Any three pupils will show a worthy result in any team olympiad. But winning the olympiad is only half the battle; first, you need to get there...

Each pupil has an independence level, expressed as an integer. In "T-generation", there is exactly one student with each independence levels from $$$l$$$ to $$$r$$$, inclusive. For a team of three pupils with independence levels $$$a$$$, $$$b$$$, and $$$c$$$, the value of their team independence is equal to $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$, where $$$\oplus$$$ denotes the bitwise XOR operation.

Your task is to choose any trio of students with the maximum possible team independence.

Input

Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case set contains two integers $$$l$$$ and $$$r$$$ ($$$0 \le l, r < 2^{30}$$$, $$$r - l > 1$$$) — the minimum and maximum independence levels of the students.

Output

For each test case set, output three pairwise distinct integers $$$a, b$$$, and $$$c$$$, such that $$$l \le a, b, c \le r$$$ and the value of the expression $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$ is maximized. If there are multiple triples with the maximum value, any of them can be output.

Example
Input
8
0 2
0 8
1 3
6 22
128 137
69 98
115 127
0 1073741823
Output
1 2 0
8 7 1
2 1 3
7 16 11
134 132 137
98 85 76
123 121 118
965321865 375544086 12551794
Note

In the first test case, the only suitable triplet of numbers ($$$a, b, c$$$) (up to permutation) is ($$$0, 1, 2$$$).

In the second test case, one of the suitable triplets is ($$$8, 7, 1$$$), where $$$(8 \oplus 7) + (7 \oplus 1) + (8 \oplus 1) = 15 + 6 + 9 = 30$$$. It can be shown that $$$30$$$ is the maximum possible value of $$$(a \oplus b) + (b \oplus c) + (a \oplus c)$$$ for $$$0 \le a, b, c \le 8$$$.