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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline", "fast-math", "unroll-loops", \
                     "no-stack-protector")
#pragma GCC push_options
#pragma GCC pop_options
#pragma GCC reset_options
#pragma GCC optimize("O3")
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastll(long long &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long logN = 17;
long long n, dfscnt;
long long rev[200005];
long long ans[200005];
long long depth[200005];
long long sparset[200005][20];
vector<long long> edge[200005];
long long fir[200005], las[200005];
long long askD(long long pos) {
  if (ans[pos] != -1) return ans[pos];
  cout << "d " << pos << '\n';
  cin >> ans[pos];
  if (ans[pos] == 0) {
    cout << "! " << pos << '\n';
    exit(0);
  }
  return ans[pos];
}
void dfs(long long pos, long long par) {
  fir[pos] = ++dfscnt;
  rev[dfscnt] = pos;
  for (int i = 0; i < edge[pos].size(); i++) {
    long long nx = edge[pos][i];
    if (nx == par) continue;
    depth[nx] = depth[pos] + 1;
    sparset[nx][0] = pos;
    dfs(nx, pos);
  }
  las[pos] = dfscnt;
}
long long lca(long long a, long long b) {
  if (depth[a] > depth[b]) swap(a, b);
  for (int i = logN; i >= 0; i--) {
    if (depth[b] - (1LL << i) < depth[a]) continue;
    b = sparset[b][i];
  }
  if (a == b) return a;
  for (int i = logN; i >= 0; i--) {
    if (sparset[a][i] == sparset[b][i]) continue;
    a = sparset[a][i];
    b = sparset[b][i];
  }
  return sparset[a][0];
}
long long dist(long long a, long long b) {
  long long curlca = lca(a, b);
  return depth[a] + depth[b] - 2 * depth[curlca];
}
vector<long long> Candidate;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  memset(ans, -1, sizeof(ans));
  memset(sparset, -1, sizeof(sparset));
  dfs(1, -1);
  for (int i = 0; i < logN; i++) {
    for (int j = 1; j <= n; j++) {
      if (sparset[j][i] == -1) continue;
      sparset[j][i + 1] = sparset[sparset[j][i]][i];
    }
  }
  long long now = askD(1);
  for (int i = 1; i <= n; i++) {
    if (depth[i] == now) Candidate.push_back(i);
  }
  bool flag = 0;
  while (Candidate.size() > 1) {
    long long u = Candidate[rng() % Candidate.size()];
    long long curD = askD(u);
    for (int i = 1; i <= curD / 2; i++) {
      u = sparset[u][0];
    }
    cout << "s " << u << '\n';
    long long root;
    cin >> root;
    now = askD(root);
    Candidate.clear();
    for (int i = fir[root]; i <= las[root]; i++) {
      long long nx = rev[i];
      if (depth[nx] - depth[root] == now) Candidate.push_back(nx);
    }
  }
  cout << "! " << Candidate[0] << '\n';
  return 0;
}