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(2)
using namespace std;
namespace Bigcat {
template <typename T>
inline void max_s(T& a, const T& b) {
  a = max(a, b);
}
template <typename T>
inline void min_s(T& a, const T& b) {
  a = min(a, b);
}
template <typename T>
void sort_v(T& x) {
  sort(x.begin(), x.end());
}
template <typename T>
void erase_dup(T& x) {
  x.erase(unique(x.begin(), x.end()), x.end());
}
template <typename T, typename V>
int pos_l(const T& x, const V& v) {
  return lower_bound(x.begin(), x.end(), v) - x.begin();
}
template <typename T, typename V>
int pos_u(const T& x, const V& v) {
  return upper_bound(x.begin(), x.end(), v) - x.begin();
}
inline unsigned rand_g() {
  static mt19937 mt_rand((int)time(0));
  return mt_rand();
}
using LL = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<LL>;
using PII = pair<int, int>;
using VPII = vector<PII>;
}  // namespace Bigcat
using namespace Bigcat;
const LL base = LL(1e9 + 7);
const LL INFL = 1LL << 62;
const int INF = 1 << 30;
const int MAX = int(2e5 + 100);
struct Node {
  int x, y, w;
};
bool operator<(const Node& n0, const Node& n1) { return n0.x < n1.y; }
int q_d(int u) {
  cout << "d " << u << endl;
  int r;
  cin >> r;
  if (r < 0) exit(1);
  return r;
}
int q_s(int u) {
  cout << "s " << u << endl;
  int r;
  cin >> r;
  if (r < 0) exit(1);
  return r;
}
void out_r(int r) {
  cout << "! " << r << endl;
  exit(0);
}
VI gr[MAX];
int st[MAX];
int dep[MAX];
int max_dep[MAX];
int candi[MAX];
int candison[MAX];
int r_dep;
void dfs(int u, int p, int d) {
  dep[u] = d;
  max_dep[u] = d;
  if (d == r_dep) {
    st[u] = 1;
    candi[u] = u;
    return;
  }
  int cs = 0;
  for (int v : gr[u]) {
    if (v != p) {
      dfs(v, u, d + 1);
      if (st[v] == 0) continue;
      st[u] += st[v];
      max_s(max_dep[u], max_dep[v]);
      if (st[v] > cs) {
        cs = st[v];
        candison[u] = v;
        candi[u] = candi[v];
      }
    }
  }
}
void work(int u, int p) {
  while (1) {
    if (dep[u] == r_dep) out_r(u);
    if (dep[u] + 1 == r_dep) {
      int r = q_s(u);
      out_r(r);
    }
    VPII zz;
    int tot = 0;
    for (int v : gr[u]) {
      if (v == p) continue;
      if (st[v] == 0) continue;
      tot += st[v];
      zz.push_back({st[v], v});
    }
    if (((int)zz.size()) == 1) {
      p = u;
      u = zz[0].second;
      continue;
    }
    sort_v(zz);
    reverse(zz.begin(), zz.end());
    int b = zz[0].second;
    if (zz[0].first > tot / 2) {
      int td = q_d(candi[b]);
      if (td == 0) out_r(candi[b]);
      int d = r_dep - td / 2;
      if (d != dep[u]) {
        p = u;
        u = b;
        while (dep[u] != d) {
          p = u;
          u = candison[u];
        }
        st[candison[u]] = 0;
        continue;
      } else {
        st[b] = 0;
        continue;
      }
    }
    {
      int t = q_s(u);
      p = u;
      u = t;
      continue;
    }
  }
}
void solve1() {
  LL result = 0;
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  r_dep = q_d(1);
  dfs(1, -1, 0);
  if (r_dep == 0) out_r(1);
  work(1, -1);
}
const bool _multi_test = false;
const int _inId = 1;
const int _outId = 0;
void io_init(char* s0) {
  string path(s0);
  int a = path.find("build\\");
  int b = path.find("Release\\");
  if (a > 0 && b > a) {
    path = path.substr(0, a) + path.substr(a + 6, b - a - 6);
  } else
    path = "";
  if (_inId > 0) {
    cerr << "redirect stdin to input " << _inId << endl;
    string s = "input";
    s += to_string(_inId);
    s += ".txt";
    if (path != "") s = path + s;
    auto r = freopen(s.c_str(), "r", stdin);
    if (r == nullptr) {
      cerr << "invalid input file " << s << endl;
      ;
      r = freopen("CON", "r", stdin);
    }
  }
  if (_outId > 0) {
    cerr << "redirect stdout to stdout " << _outId << endl;
    string s = "stdout";
    s += to_string(_outId);
    s += ".txt";
    if (path != "") s = path + s;
    auto r = freopen(s.c_str(), "w", stdout);
    if (r == nullptr) {
      cerr << "invalid output file " << s << endl;
      ;
      r = freopen("CON", "w", stdout);
    }
  }
}
const bool _local_test = false;
void solve() {
  int t = 1;
  if (_multi_test) cin >> t;
  if (_local_test) cerr << " now base " << base << endl;
  for (int i = 0; i < t; ++i) {
    if (_local_test) cerr << " tasks: " << i + 1 << endl;
    solve1();
  }
  int x = 1;
}
signed main(signed argc, char* argv[]) {
  if (_local_test) {
    if (argc == 1) {
      io_init(argv[0]);
    }
  }
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  cout.precision(11);
  solve();
}