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<int> nei[1000000];
int size[1000000];
int dist[1000000];
vector<int> child[1000000];
int parent[1000000];
int prefered[1000000];
void DFS(int node) {
  size[node] = 1;
  prefered[node] = -1;
  for (auto v : nei[node]) {
    if (size[v] == -1) {
      child[node].push_back(v);
      dist[v] = dist[node] + 1;
      parent[v] = node;
      DFS(v);
      size[node] += size[v];
    }
  }
  for (auto v : nei[node]) {
    if (dist[v] > dist[node] && size[v] * 2 >= size[node]) {
      prefered[node] = v;
    }
  }
}
int hidden;
int queries;
void query(int node) {
  if (dist[node] == hidden) {
    cout << "! " << node + 1 << endl;
    return;
  }
  int Q = node;
  while (prefered[Q] != -1) {
    Q = prefered[Q];
  }
  cout << "d " << Q + 1 << endl;
  int ans;
  cin >> ans;
  int anc = (ans + dist[Q] - hidden);
  anc /= 2;
  for (int i = 0; i < anc; i++) {
    Q = parent[Q];
  }
  if (dist[Q] == hidden) {
    cout << "! " << Q + 1 << endl;
    return;
  }
  cout << "s " << Q + 1 << endl;
  cin >> Q;
  Q--;
  query(Q);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    nei[x].push_back(y);
    nei[y].push_back(x);
  }
  for (int i = 0; i < n; i++) size[i] = -1;
  dist[0] = 0;
  DFS(0);
  cout << "d 1" << endl;
  cin >> hidden;
  queries = 1;
  query(0);
  return 0;
}