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 long long MAX_N = 2e5 + 30;
long long n, d, V;
long long hi[MAX_N];
long long down[MAX_N];
long long par[MAX_N];
vector<int> G[MAX_N];
set<pair<int, int> > Set;
set<pair<int, int> >::iterator it;
int dfs(int v = 0, int p = 0, int h = 0) {
  par[v] = p;
  hi[v] = h;
  if (h == d) {
    down[v] = 1;
    return 1;
  }
  for (int u : G[v])
    if (u != p) down[v] += dfs(u, v, h + 1);
  return down[v];
}
pair<int, int> get() {
  if (!Set.size()) return {-1, -1};
  it = Set.end();
  it--;
  return *it;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  cout << "d " << 1 << endl << flush;
  cin >> d;
  dfs();
  V = 0;
  while (hi[V] != d) {
    int sum = 0, mx = 0;
    for (int u : G[V])
      if (u != par[V]) {
        sum += down[u];
        if (mx < down[u]) mx = down[u];
      }
    if (mx * 2 <= sum) {
      cout << "s " << V + 1 << endl << flush;
      int tmp;
      cin >> tmp;
      V = tmp - 1;
      continue;
    }
    Set.clear();
    sum = 0;
    for (int u : G[V])
      if (u != par[V]) {
        sum += down[u];
        Set.insert({down[u], u});
      }
    int U = V;
    while (sum < get().first * 2) {
      int v = get().second;
      U = v;
      sum -= down[v];
      Set.erase(get());
      for (int u : G[v])
        if (u != par[v]) {
          sum += down[u];
          Set.insert({down[u], u});
        }
    }
    if (Set.size() <= 1) {
      V = U;
      break;
    }
    cout << "d " << U + 1 << endl << flush;
    int len;
    cin >> len;
    while (hi[U] + len > d) {
      U = par[U];
      len--;
    }
    if (hi[U] == d) {
      V = U;
      break;
    }
    cout << "s " << U + 1 << endl << flush;
    int tmp;
    cin >> tmp;
    V = tmp - 1;
    Set.clear();
  }
  cout << "! " << V + 1 << endl << flush;
  return 0;
}