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>
using namespace std;
vector<vector<int> > neigh;
vector<int> subtree_size, best_index, best_neigh, dist;
void DFS(int a, int par) {
  subtree_size[a] = 1;
  best_index[a] = a;
  best_neigh[a] = -1;
  for (int b : neigh[a])
    if (b != par) {
      dist[b] = dist[a] + 1;
      DFS(b, a);
      subtree_size[a] += subtree_size[b];
      if (best_neigh[a] == -1 ||
          subtree_size[best_neigh[a]] < subtree_size[b]) {
        best_index[a] = best_index[b];
        best_neigh[a] = b;
      }
    }
}
int Ask_dist(int a) {
  cout << "d " << a + 1 << endl;
  int dist;
  cin >> dist;
  return dist;
}
int Ask_second(int a) {
  cout << "s " << a + 1 << endl;
  int second;
  cin >> second;
  return second - 1;
}
void Answer(int a) { cout << "! " << a + 1 << endl; }
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  neigh.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    neigh[a].push_back(b);
    neigh[b].push_back(a);
  }
  subtree_size.resize(n);
  best_index.resize(n);
  best_neigh.resize(n);
  dist.resize(n);
  dist[0] = 0;
  DFS(0, -1);
  int cur_i = 0;
  int cur_dist = Ask_dist(cur_i);
  while (cur_dist > 0) {
    cerr << cur_i << ' ' << cur_dist << endl;
    if (cur_dist == 1 ||
        subtree_size[best_neigh[cur_i]] * 2 < subtree_size[cur_i] - 1) {
      cur_i = Ask_second(cur_i);
      cur_dist--;
    } else {
      int target_dist = (Ask_dist(best_index[cur_i]) + cur_dist + dist[cur_i] -
                         dist[best_index[cur_i]]) /
                        2;
      while (cur_dist > target_dist) {
        cur_i = best_neigh[cur_i];
        cur_dist--;
      }
      subtree_size[cur_i] -= subtree_size[best_neigh[cur_i]];
      int new_best_neigh = -1;
      for (int b : neigh[cur_i])
        if (b != best_neigh[cur_i] && subtree_size[b] < subtree_size[cur_i] &&
            (new_best_neigh == -1 ||
             subtree_size[b] > subtree_size[new_best_neigh])) {
          new_best_neigh = b;
        }
      best_neigh[cur_i] = new_best_neigh;
      best_index[cur_i] = best_index[new_best_neigh];
    }
  }
  Answer(cur_i);
}