#P1305D. Kuroni and the Celebration

    ID: 1867 远端评测题 1000ms 256MiB 尝试: 0 已通过: 0 难度: (无) 上传者: 标签>constructive algorithmsdfs and similarinteractivetrees*1900

Kuroni and the Celebration

Description

This is an interactive problem.

After getting AC after 13 Time Limit Exceeded verdicts on a geometry problem, Kuroni went to an Italian restaurant to celebrate this holy achievement. Unfortunately, the excess sauce disoriented him, and he's now lost!

The United States of America can be modeled as a tree (why though) with nn vertices. The tree is rooted at vertex rr, wherein lies Kuroni's hotel.

Kuroni has a phone app designed to help him in such emergency cases. To use the app, he has to input two vertices uu and vv, and it'll return a vertex ww, which is the lowest common ancestor of those two vertices.

However, since the phone's battery has been almost drained out from live-streaming Kuroni's celebration party, he could only use the app at most n2\lfloor \frac{n}{2} \rfloor times. After that, the phone would die and there will be nothing left to help our dear friend! :(

As the night is cold and dark, Kuroni needs to get back, so that he can reunite with his comfy bed and pillow(s). Can you help him figure out his hotel's location?

Interaction

The interaction starts with reading a single integer nn (2n10002 \le n \le 1000), the number of vertices of the tree.

Then you will read n1n-1 lines, the ii-th of them has two integers xix_i and yiy_i (1xi,yin1 \le x_i, y_i \le n, xiyix_i \ne y_i), denoting there is an edge connecting vertices xix_i and yiy_i. It is guaranteed that the edges will form a tree.

Then you can make queries of type "? u v" (1u,vn1 \le u, v \le n) to find the lowest common ancestor of vertex uu and vv.

After the query, read the result ww as an integer.

In case your query is invalid or you asked more than n2\lfloor \frac{n}{2} \rfloor queries, the program will print 1-1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you find out the vertex rr, print "! rr" and quit after that. This query does not count towards the n2\lfloor \frac{n}{2} \rfloor limit.

Note that the tree is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.

After printing any query do not forget to print end of line and flush the output. Otherwise, you might 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.

Hacks

To hack, use the following format:

The first line should contain two integers nn and rr (2n10002 \le n \le 1000, 1rn1 \le r \le n), denoting the number of vertices and the vertex with Kuroni's hotel.

The ii-th of the next n1n-1 lines should contain two integers xix_i and yiy_i (1xi,yin1 \le x_i, y_i \le n) — denoting there is an edge connecting vertex xix_i and yiy_i.

The edges presented should form a tree.

Samples

样例输入 1

6
1 4
4 2
5 3
6 3
2 3

3

4

4

样例输出 1

? 5 6

? 3 1

? 1 2

! 4

Note

Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.

The image below demonstrates the tree in the sample test: