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 nax = 2e5 + 5;
vector<int> adj[nax];
int t = 1;
vector<int> preorder(nax, 0);
vector<int> postorder(nax, 0);
vector<bool> blocked(nax, false);
vector<int> poziom(nax, 0);
vector<int> poddrzewo(nax, 0);
vector<int> ojciec(nax, -1);
int poziom_x;
int n;
int akt = 1;
int roz;
void dfs(int v, int p) {
  preorder[v] = t++;
  ojciec[v] = p;
  if (v == 1)
    poziom[v] = 0;
  else
    poziom[v] = poziom[p] + 1;
  for (int i = 0; i < adj[v].size(); i++) {
    if (adj[v][i] == p) continue;
    dfs(adj[v][i], v);
  }
  postorder[v] = t++;
}
bool czy_ancestor(int u, int v) {
  if (preorder[u] < preorder[v] && postorder[u] > postorder[v]) return true;
  return false;
}
int przelicz(int v, int p) {
  poddrzewo[v] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    if (blocked[adj[v][i]] == true || adj[v][i] == p) continue;
    poddrzewo[v] += przelicz(adj[v][i], v);
  }
  return poddrzewo[v];
}
int znajdz_centroid(int v) {
  przelicz(v, v);
  int p = -1;
  while (1) {
    bool ok = true;
    for (int i = 0; i < adj[v].size(); i++) {
      if (blocked[adj[v][i]] == true || adj[v][i] == p) continue;
      int syn = adj[v][i];
      if (poddrzewo[syn] * 2 > roz) {
        p = v;
        v = syn;
        ok = false;
        break;
      }
    }
    if (ok) {
      return v;
    }
  }
}
int d(int x) {
  cout << 'd' << " " << x << endl;
  fflush(stdout);
  int xd;
  cin >> xd;
  return xd;
}
int s(int x) {
  cout << 's' << " " << x << endl;
  fflush(stdout);
  int xd;
  cin >> xd;
  return xd;
}
void wypisz(int x) {
  cout << "!"
       << " " << x << endl;
  fflush(stdout);
  exit(0);
}
int ilema(int v, int p) {
  int res = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    if (blocked[adj[v][i]] == true || adj[v][i] == p) continue;
    res += ilema(adj[v][i], v);
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 1);
  roz = n;
  poziom_x = d(1);
  while (roz > 2) {
    int v = znajdz_centroid(akt);
    int odl = d(v);
    if (odl == 0) wypisz(v);
    blocked[v] = true;
    if (poziom_x != odl + poziom[v]) {
      akt = ojciec[v];
      roz = ilema(akt, akt);
    } else {
      akt = s(v);
      roz = ilema(akt, akt);
    }
  }
  if (roz == 1)
    wypisz(akt);
  else {
    int dist = d(akt);
    if (dist == 0) wypisz(akt);
    for (int i = 0; i < adj[akt].size(); i++) {
      if (blocked[adj[akt][i]] == false) wypisz(adj[akt][i]);
    }
  }
  return 0;
}