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;
int n;
int cur_node, cur_level;
int x_level;
vector<set<int>> edges;
map<int, int> parent;
vector<int> subtree_size;
void Load() {
  scanf("%d", &n);
  edges.resize(n + 1);
  subtree_size.resize(n + 1);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    edges[u].insert(v);
    edges[v].insert(u);
  }
}
int AskDistanceFrom(int u) {
  printf("d %d\n", u);
  fflush(stdout);
  int dist;
  scanf("%d", &dist);
  return dist;
}
int AskSecondNodeToXFrom(int u) {
  printf("s %d\n", u);
  fflush(stdout);
  int ans;
  scanf("%d", &ans);
  return ans;
}
void PrintAnswerAndExit(int x) {
  printf("! %d\n", x);
  fflush(stdout);
  exit(0);
}
int PopulateTreeDataDFS(int u, int par) {
  subtree_size[u] = 1;
  for (int v : edges[u]) {
    if (v == par) continue;
    parent[v] = u;
    subtree_size[u] += PopulateTreeDataDFS(v, u);
  }
  return subtree_size[u];
}
void DeleteSubtreeOf(int u) {
  edges[parent[u]].erase(u);
  int cur = u;
  while (cur != 1) {
    subtree_size[parent[cur]] -= subtree_size[u];
    cur = parent[cur];
  }
}
bool AnyChildHasMajority(int u) {
  for (int v : edges[u]) {
    if (v == parent[u]) continue;
    if (subtree_size[v] * 2 > subtree_size[u]) return true;
  }
  return false;
}
std::tuple<int, int> FindSmallestRightDescendantWithMajority(int u) {
  int curr = u, dist = 0;
  while (true) {
    bool found = false;
    for (int v : edges[curr]) {
      if (v == parent[curr]) continue;
      if (subtree_size[v] * 2 > subtree_size[u]) {
        found = true;
        curr = v;
        dist++;
        break;
      }
    }
    if (!found) break;
  }
  return {curr, dist};
}
void Solve() {
  cur_node = 1, cur_level = 0;
  x_level = AskDistanceFrom(1);
  if (x_level == 0) {
    PrintAnswerAndExit(1);
  }
  PopulateTreeDataDFS(1, -1);
  while (cur_level != x_level) {
    if (!AnyChildHasMajority(cur_node)) {
      cur_node = AskSecondNodeToXFrom(cur_node);
      cur_level++;
      continue;
    }
    int rj, dist_cur_to_rj;
    std::tie(rj, dist_cur_to_rj) =
        FindSmallestRightDescendantWithMajority(cur_node);
    int dist_from_rj = AskDistanceFrom(rj);
    if (dist_cur_to_rj + dist_from_rj == x_level - cur_level) {
      cur_node = rj;
      cur_level = x_level - dist_from_rj;
      continue;
    }
    DeleteSubtreeOf(rj);
  }
  PrintAnswerAndExit(cur_node);
}
int main() {
  Load();
  Solve();
}