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;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    if (i) os << ", ";
    os << v[i];
  }
  return os << "]";
}
void debug() { cerr << "\n"; }
template <typename T, typename... Args>
void debug(T x, Args... args) {
  cerr << x << "; ";
  debug(args...);
}
int depthX, depth[200000], sub[200000];
vector<int> path, adj[200000];
int query(char c, int u) {
  cout << c << " " << u + 1 << endl;
  int ret;
  cin >> ret;
  return ret;
}
int dfsSub(int u, int p) {
  sub[u] = 1;
  for (int v : adj[u])
    if (v != p) {
      depth[v] = depth[u] + 1;
      sub[u] += dfsSub(v, u);
    }
  return sub[u];
}
void dfsHld(int u, int p) {
  path.push_back(u);
  int nxt = -1;
  for (int v : adj[u])
    if (v != p && (nxt == -1 || sub[v] > sub[nxt])) nxt = v;
  if (nxt != -1) dfsHld(nxt, u);
}
int dfs(int u, int p) {
  path.clear();
  dfsHld(u, p);
  int depthY = (depthX + depth[path.back()] - query('d', path.back())) / 2,
      y = path[depthY - depth[u]];
  if (depthX == depthY) return y;
  return dfs(query('s', y) - 1, y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u - 1].push_back(v - 1);
    adj[v - 1].push_back(u - 1);
  }
  depthX = query('d', 0);
  depth[0] = 0;
  dfsSub(0, -1);
  cout << "! " << dfs(0, -1) + 1 << endl;
  return 0;
}