Evirir the dragon has many friends. They have 3 friends! That is one more than the average dragon.

You are given integers $n$, $x$, and $y$. There are $n$ dragons sitting in a circle. The dragons are numbered $1, 2, \ldots, n$. For each $i$ ($1 \le i \le n$), dragon $i$ is friends with dragon $i - 1$ and $i + 1$, where dragon $0$ is defined to be dragon $n$ and dragon $n + 1$ is defined to be dragon $1$. Additionally, dragons $x$ and $y$ are friends with each other (if they are already friends, this changes nothing). Note that all friendships are mutual.

Output $n$ non-negative integers $a_1, a_2, \ldots, a_n$ such that for each dragon $i$ ($1 \le i \le n$), the following holds:

• Let $f_1, f_2, \ldots, f_k$ be the friends of dragon $i$. Then $a_i = \operatorname{mex}(a_{f_1}, a_{f_2}, \ldots, a_{f_k})$.$^{\text{∗}}$

$^{\text{∗}}$The minimum excluded (MEX) of a collection of integers $c_1, c_2, \ldots, c_m$ is defined as the smallest non-negative integer $t$ which does not occur in the collection $c$.

Input

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

The first and only line of each test case contains three integers $n$, $x$, $y$ ($3 \le n \le 2 \cdot 10^5$, $1 \le x \lt y \le n$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

Output

For each test case, output $n$ space-separated non-negative integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 10^9$) on a line that satisfy the condition in the statement. If there are multiple solutions, print any of them. It can be proven that under the problem constraints, a solution with $0 \le a_i \le 10^9$ always exists.

Note

For the first test case:

• $i = 1$: Dragon $1$'s friends are dragons $2, 3, 5$. $\operatorname{mex}(a_2, a_3, a_5) = \operatorname{mex}(2, 1, 1) = 0 = a_1$, so the condition for dragon $1$ is satisfied.
• $i = 2$: Dragon $2$'s friends are dragons $1, 3$. $\operatorname{mex}(a_1, a_3) = \operatorname{mex}(0, 1) = 2 = a_2$.
• $i = 3$: Dragon $3$'s friends are dragons $1, 2, 4$. $\operatorname{mex}(a_1, a_2, a_4) = \operatorname{mex}(0, 2, 0) = 1 = a_3$.
• $i = 4$: Dragon $4$'s friends are dragons $3, 5$. $\operatorname{mex}(a_3, a_5) = \operatorname{mex}(1, 1) = 0 = a_4$.
• $i = 5$: Dragon $5$'s friends are dragons $1, 4$. $\operatorname{mex}(a_1, a_4) = \operatorname{mex}(0, 0) = 1 = a_5$.