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;
const int maxn = 2e5 + 42;
int n;
vector<int> graph[maxn];
int size_[maxn];
int heavy[maxn];
int h[maxn];
int parent[maxn];
void dfs(int v) {
  size_[v] = 1;
  int maxchild = 0;
  for (auto to : graph[v]) {
    if (to != parent[v]) {
      parent[to] = v;
      h[to] = h[v] + 1;
      dfs(to);
      size_[v] += size_[to];
      if (size_[to] > maxchild) {
        maxchild = size_[to];
        heavy[v] = to;
      }
    }
  }
}
int askD(int v) {
  cout << "d " << v << endl;
  int x;
  cin >> x;
  assert(x != -1);
  return x;
}
int askS(int v) {
  cout << "s " << v << endl;
  int x;
  cin >> x;
  assert(x != -1);
  return x;
}
void ok(int v) {
  cout << "! " << v << endl;
  exit(0);
}
void solve(int v, int depth) {
  if (depth == 0) {
    ok(v);
  }
  vector<int> path;
  for (int i = v; i != 0; i = heavy[i]) {
    path.push_back(i);
  }
  int xleaf = askD(path.back());
  int vleaf = h[path.back()] - h[v];
  int a = ((depth - h[v]) + vleaf - xleaf);
  assert(a >= 0 && a % 2 == 0);
  a /= 2;
  assert(a < ((int)(path).size()));
  int u = path[a];
  if (a == depth - h[v]) {
    ok(u);
  }
  solve(askS(u), depth);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  dfs(1);
  int depth = askD(1);
  solve(1, depth);
}