B. Sum Graph
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

This is an interactive problem.

There is a hidden permutation $p_1, p_2, \dots, p_n$.

Consider an undirected graph with $n$ nodes only with no edges. You can make two types of queries:

1. Specify an integer $x$ satisfying $2 \le x \le 2n$. For all integers $i$ ($1 \le i \le n$) such that $1 \le x-i \le n$, an edge between node $i$ and node $x-i$ will be added.
2. Query the number of edges in the shortest path between node $p_i$ and node $p_j$. As the answer to this question you will get the number of edges in the shortest path if such a path exists, or $-1$ if there is no such path.

Note that you can make both types of queries in any order.

Within $2n$ queries (including type $1$ and type $2$), guess two possible permutations, at least one of which is $p_1, p_2, \dots, p_n$. You get accepted if at least one of the permutations is correct. You are allowed to guess the same permutation twice.

A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).

Input

Each test contains multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \le n \le 10^3$) — the length of the permutation.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^3$.

Interaction

The interaction for each test case begins by reading the integer $n$.

Then, make at most $2n$ queries:

• If you want to make a type $1$ query, output "+ x". $x$ must be an integer between $2$ and $2n$ inclusive. After doing that read $1$ or $-2$. If you read $1$ your query was valid, otherwise it was invalid or you exceed the limit of queries, and your program must terminate immediately to receive a Wrong Answer verdict.
• If you want to make a type $2$ query, output "? i j". $i$ and $j$ must be integers between $1$ and $n$ inclusive. After that, read in a single integer $r$ ($-1 \le r \le n$) — the answer to your query. If you receive the integer $−2$ instead of an answer, it means your program has made an invalid query, or has exceeded the limit of queries. Your program must terminate immediately to receive a Wrong Answer verdict.

At any point of the interaction, if you want to guess two permutations, output "! $p_{1,1}$ $p_{1,2}$ $\dots$ $p_{1,n}$ $p_{2,1}$ $p_{2,2}$ $\dots$ $p_{2,n}$". Note that you should output the two permutations on the same line, and no exclamation mark is needed to separate the two permutations. After doing that read $1$ or $-2$. If you read $1$ your answer was correct, otherwise it was incorrect and your program must terminate immediately to receive a Wrong Answer verdict. After that, move on to the next test case, or terminate the program if there are none. Note that reporting the answer does not count as a query.

Note that even if you output a correct permutation, the second permutation should be a permutation and not an arbitrary array.

At any point, if you continue interaction after reading in the integer $-2$, you can get an arbitrary verdict because your solution will continue to read from a closed stream.

After printing a query or the answer do not forget to output the end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

• fflush(stdout) or cout.flush() in C++;
• System.out.flush() in Java;
• flush(output) in Pascal;
• stdout.flush() in Python;
• see the documentation for other languages.

Interactor is non-adaptive. This means that all permutations are fixed before the interaction starts.

Hacks

To make a hack, use the following format.

The first line should contain a single integer $t$ ($1 \le t \le 100$) — the number of test cases.

The first line of each test case should contain a single integer $n$ ($2 \le n \le 10^3$) — the length of the permutation.

The second line of each test case should contain $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$) — the hidden permutation.

The sum of $n$ over all test cases should not exceed $10^3$.

Example
Input
2
6

1

1

1

1

1

2

-1

1
2

1

Output

+ 12

+ 2

+ 3

? 1 3

+ 5

? 1 5

? 4 5

! 1 4 2 5 3 6 1 2 3 4 5 6

! 1 2 2 1


Note

In the first test case, $n=6$ and the hidden permutation $p = [1,4,2,5,3,6]$.

Firstly, make a type $1$ query on $x=12, 2, 3$ respectively. This adds four edges to the graph in total:

• An edge that connects node $6$ and node $6$.
• An edge that connects node $1$ and node $1$.
• An edge that connects node $1$ and node $2$.
• An edge that connects node $2$ and node $1$.

Since all of these queries are valid, the interactor returns $1$ after each of them.

Then, query the number of edges in the shortest path between node $p_1 = 1$ and $p_3 = 2$, which is equal to $1$.

Then, make a type $1$ query on $x=5$. This adds four edges to the graph in total:

• An edge that connects node $1$ and node $4$.
• An edge that connects node $2$ and node $3$.
• An edge that connects node $3$ and node $2$.
• An edge that connects node $4$ and node $1$.

Since this query is valid, the interactor returns $1$.

Then, query the number of edges in the shortest path between node $p_1 = 1$ and $p_5 = 3$, which is equal to $2$.

Then, query the number of edges in the shortest path between node $p_4 = 5$ and $p_5 = 3$. Such a path doesn't exist, therefore the interactor returns $-1$.

Afterwards, due to some magic, two possible permutations that can be $p$ are determined: the first permutation is $[1,4,2,5,3,6]$ and the second permutation is $[1,2,3,4,5,6]$. Since the first permutation is equal to the hidden permutation, this test case is solved correctly. In total, $7$ queries are used, which is within the limit of $2 \cdot 6 = 12$ queries.

Since the answer is correct, the interactor returns $1$.

In the second test case, $n=2$ and the hidden permutation is $p = [2,1]$.

Since there are only $2! = 2$ possible permutations, no queries are needed. It is sufficient to just output the two permutations, $[1,2]$ and $[2,1]$. In total, $0$ queries are used, which is within the limit of $2 \cdot 2 = 4$ queries.

Since the answer is correct, the interactor returns $1$.