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 <class T, class U>
inline bool chkmax(T& a, U b) {
  return (a < b) ? (a = b, true) : false;
}
template <class T, class U>
inline bool chkmin(T& a, U b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline vector<T> uniq(vector<T> vec) {
  sort((vec).begin(), (vec).end()),
      vec.erase(unique((vec).begin(), (vec).end()), end(vec));
  return vec;
}
void solve();
int32_t main(int argc, char** argv) {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T = 1;
  while (T--) solve();
  double times = (clock() * 1.0 / CLOCKS_PER_SEC);
  do {
  } while (0);
}
int n, rtDis;
vector<vector<int> > G;
vector<int> size, son, dep;
void init(int n) { G.resize(n), size.resize(n), son.resize(n), dep.resize(n); }
int query(char opt, int vertex) {
  (cout << opt << " " << vertex + 1 << "\n").flush();
  int x;
  cin >> x;
  if (opt == 's') --x;
  return x;
}
void answer(int x) {
  (cout << "! " << x + 1 << "\n").flush();
  exit(0);
}
void dfs(int u, int f, int d) {
  size[u] = 1, son[u] = -1, dep[u] = d;
  for (auto& v : G[u]) {
    if (v != f) {
      dfs(v, u, d + 1);
      size[u] += size[v];
      if (son[u] == -1 || size[v] > size[son[u]]) son[u] = v;
    }
  }
}
int getans(int u) {
  vector<int> heavy;
  for (int i = u; i != -1; i = son[i]) {
    heavy.push_back(i);
  }
  int endpos = heavy.back(), dis = query('d', endpos);
  int pos = (rtDis + dep[endpos] - dis) / 2 - dep[u], y = heavy[pos];
  if (dep[y] == rtDis) return y;
  return getans(query('s', y));
}
void solve() {
  cin >> n;
  init(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);
  }
  dfs(0, -1, 0);
  rtDis = query('d', 0);
  answer(getans(0));
}