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>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
void err(istream_iterator<string>) {}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> buf) {
  return os << "(" << buf.first << ": " << buf.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, vector<vector<T>> &A) {
  for (auto &B : A) os << '\n' << B;
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &A) {
  bool f = false;
  os << "{";
  for (const auto e : A) {
    if (f) os << ", ";
    os << e;
    f = true;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, set<T> &A) {
  bool f = false;
  os << "{";
  for (const auto e : A) {
    if (f) os << ", ";
    os << e;
    f = true;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> &A) {
  bool f = false;
  os << "{";
  for (const auto e : A) {
    if (f) os << ", ";
    os << e;
    f = true;
  }
  return os << "}";
}
template <typename K, typename V>
ostream &operator<<(ostream &os, map<K, V> &A) {
  bool f = false;
  os << "{";
  for (auto &e : A) {
    if (f) os << ", ";
    os << e;
    f = true;
  }
  return os << "}";
}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << a << endl;
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 1000000007;
long long const INF = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int randint(int a, int b) {
  return uniform_int_distribution<int>(a, b)(rng);
}
int dist(int u) {
  cout << "d " << u + 1 << endl;
  int ret;
  cin >> ret;
  return ret;
}
int go(int u) {
  cout << "s " << u + 1 << endl;
  int ret;
  cin >> ret;
  return --ret;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int depth = dist(0);
  vector<int> a, par(n, -1);
  function<void(int, int)> dfs = [&](int u, int d) {
    if (d == depth) {
      a.push_back(u);
      return;
    }
    for (int v : g[u])
      if (v != par[u]) {
        par[v] = u;
        dfs(v, 1 + d);
      }
  };
  dfs(0, 0);
  while (int((a).size()) > 1) {
    int u = a[randint(0, int((a).size()) - 1)];
    int p = dist(u) / 2;
    if (p == 0) {
      a = {u};
      break;
    }
    depth = p - 1;
    while (p--) u = par[u];
    u = go(u);
    a = {};
    dfs(u, 0);
  }
  cout << "! " << a[0] + 1 << endl;
  return 0;
}