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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
vector<vector<int>> gr;
vector<int> par, d;
void dfs_parents(int v, int p = -1) {
  if (p != -1) par[v] = p, d[v] = d[p] + 1;
  for (int u : gr[v])
    if (u != p) dfs_parents(u, v);
}
vector<int> sz;
vector<bool> is_deleted;
int recalc_sz(int v, int p) {
  sz[v] = 1;
  for (int u : gr[v])
    if (u != p && !is_deleted[u]) sz[v] += recalc_sz(u, v);
  return sz[v];
}
int find_centroid(int v, int p, int subtree_sz) {
  for (int u : gr[v])
    if (u != p && !is_deleted[u] && 2 * sz[u] >= subtree_sz)
      return find_centroid(u, v, subtree_sz);
  return v;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  gr.resize(n);
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u, v--, u--;
    gr[v].push_back(u), gr[u].push_back(v);
  }
  par = d = sz = vector<int>(n);
  is_deleted = vector<bool>(n);
  dfs_parents(0);
  int v = 0, depth;
  cout << "d 1" << endl;
  cin >> depth;
  while (true) {
    recalc_sz(v, -1);
    if (sz[v] == 1) return cout << "! " << v + 1 << endl, 0;
    v = find_centroid(v, -1, sz[v]);
    is_deleted[v] = 1;
    cout << "d " << v + 1 << endl;
    int ans;
    cin >> ans;
    if (!ans) return cout << "! " << v + 1 << endl, 0;
    if (d[v] + ans != depth) {
      v = par[v];
    } else {
      cout << "s " << v + 1 << endl;
      int u;
      cin >> u, u--;
      v = u;
    }
  }
  return 0;
}