Problem: This is an interactive problem.

You're given a tree consisting of n nodes, rooted at node 1. A tree is a connected graph with no cycles.

We chose a hidden node x. In order to find this node, you can ask queries of two types: 

  * d u (1 ≤ u ≤ n). We will answer with the distance between nodes u and x. The distance between two nodes is the number of edges in the shortest path between them. 
  * s u (1 ≤ u ≤ n). We will answer with the second node on the path from u to x. However, there's a plot twist. If u is not an ancestor of x, you'll receive "Wrong answer" verdict! 



Node a is called an ancestor of node b if a ≠ b and the shortest path from node 1 to node b passes through node a. Note that in this problem a node is not an ancestor of itself.

Can you find x in no more than 36 queries? The hidden node is fixed in each test beforehand and does not depend on your queries.

Input

The first line contains the integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines contains two space-separated integers u and v (1 ≤ u,v ≤ n) that mean there's an edge between nodes u and v. It's guaranteed that the given graph is a tree.

Output

To print the answer, print "! x" (without quotes).

Interaction

To ask a question, print it in one of the formats above: 

  * d u (1 ≤ u ≤ n), or 
  * s u (1 ≤ u ≤ n). 



After each question, you should read the answer: either the distance or the second vertex on the path, as mentioned in the legend. 

If we answer with -1 instead of a valid answer, that means you exceeded the number of queries, made an invalid query, or violated the condition in the second type of queries. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise, you can get an arbitrary verdict because your solution will continue to read from a closed stream.

After printing a query, do not forget to output 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.



Hacks:

The first line should contain two integers n and x (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ x ≤ n).

Each of the next n-1 lines should contain two integers u and v (1 ≤ u,v ≤ n) that mean there is an edge between nodes u and v. The edges must form a tree.

Example

Input


5
1 2
1 3
3 4
3 5
3
5

Output


d 2
s 3
! 5

Note

In the first example, the hidden node is node 5.

<image>

We first ask about the distance between node x and node 2. The answer is 3, so node x is either 4 or 5. We then ask about the second node in the path from node 3 to node x. Note here that node 3 is an ancestor of node 5. We receive node 5 as the answer. Finally, we report that the hidden node is node 5.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
void dfs(int32_t vertex, int32_t parent, std::vector<int32_t>* tree,
         int32_t* subtree_sizes, int32_t* depths, int32_t depth,
         int32_t* parents) {
  subtree_sizes[vertex] = 1;
  depths[vertex] = depth;
  parents[vertex] = parent;
  for (int32_t i = 0; i < tree[vertex].size(); i++)
    if (tree[vertex][i] != parent) {
      dfs(tree[vertex][i], vertex, tree, subtree_sizes, depths, depth + 1,
          parents);
      subtree_sizes[vertex] += subtree_sizes[tree[vertex][i]];
    }
}
int32_t next_path = 1;
void build_hld(int32_t vertex, int32_t parent, std::vector<int32_t>* tree,
               int32_t* path_index, int32_t* lowest_per_path, int32_t cur_path,
               int32_t* subtree_sizes) {
  path_index[vertex] = cur_path;
  lowest_per_path[cur_path] = vertex;
  std::vector<int32_t> children;
  for (int32_t i = 0; i < tree[vertex].size(); i++)
    if (tree[vertex][i] != parent) children.push_back(tree[vertex][i]);
  if (children.size() == 0) return;
  int32_t fattest_child = 0;
  for (int32_t i = 0; i < children.size(); i++)
    if (subtree_sizes[children[i]] > subtree_sizes[children[fattest_child]])
      fattest_child = i;
  if (subtree_sizes[children[fattest_child]] * 2 >= subtree_sizes[vertex])
    build_hld(children[fattest_child], vertex, tree, path_index,
              lowest_per_path, cur_path, subtree_sizes);
  else
    build_hld(children[fattest_child], vertex, tree, path_index,
              lowest_per_path, next_path++, subtree_sizes);
  for (int32_t i = 0; i < children.size(); i++)
    if (i != fattest_child)
      build_hld(children[i], vertex, tree, path_index, lowest_per_path,
                next_path++, subtree_sizes);
}
int32_t kth_ancestor(int32_t vertex, int32_t k, int32_t* parents) {
  int32_t cur = vertex;
  for (int32_t i = 0; i < k; i++) cur = parents[cur];
  return cur;
}
int main() {
  int32_t n;
  std::cin >> n;
  std::vector<int32_t>* tree = new std::vector<int32_t>[n];
  for (int32_t i = 0; i < n - 1; i++) {
    int32_t src, dst;
    std::cin >> src >> dst;
    src--;
    dst--;
    tree[src].push_back(dst);
    tree[dst].push_back(src);
  }
  int32_t* subtree_sizes = new int32_t[n];
  int32_t* depths = new int32_t[n];
  int32_t* parents = new int32_t[n];
  dfs(0, -1, tree, subtree_sizes, depths, 0, parents);
  int32_t* path_index = new int32_t[n];
  int32_t* lowest_per_path = new int32_t[n];
  build_hld(0, -1, tree, path_index, lowest_per_path, 0, subtree_sizes);
  int32_t dist_from_1;
  std::cout << "d 1" << std::endl;
  std::cin >> dist_from_1;
  int32_t cur = 0;
  while (true) {
    int32_t dist_to_x;
    std::cout << "d " << lowest_per_path[path_index[cur]] + 1 << std::endl;
    std::cin >> dist_to_x;
    int32_t difference =
        (dist_from_1 + dist_to_x - depths[lowest_per_path[path_index[cur]]]) /
        2;
    int32_t last_here = kth_ancestor(lowest_per_path[path_index[cur]],
                                     dist_to_x - difference, parents);
    if (difference == 0) {
      std::cout << "! " << last_here + 1 << std::endl;
      return 0;
    }
    std::cout << "s " << last_here + 1 << " ";
    std::cin >> cur;
    cur--;
  }
}