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 Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
constexpr int MAXN = 200000;
vector<int> gr[MAXN];
bool cand[MAXN];
int depth[MAXN];
int cnt[MAXN];
int prnt[MAXN];
inline int AskDist(int v) {
  cout << "d " << v + 1 << "\n" << flush;
  int res;
  cin >> res;
  assert(res != -1);
  return res;
}
inline int AskNext(int v) {
  cout << "s " << v + 1 << "\n" << flush;
  int res;
  cin >> res;
  assert(res != -1);
  return res - 1;
}
inline void Answer(int v) { cout << "! " << v + 1 << "\n" << flush; }
void Dfs(int h, int from, int p = -1) {
  prnt[from] = p;
  cand[from] = depth[from] == h;
  cnt[from] = cand[from];
  for (int to : gr[from]) {
    if (to != p) {
      depth[to] = depth[from] + 1;
      Dfs(h, to, from);
      cnt[from] += cnt[to];
    }
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < (n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  int dist = AskDist(0);
  Dfs(dist, 0);
  for (int from = 0;;) {
    if (dist == 0) {
      Answer(from);
      break;
    }
    int mid = cnt[from] >> 1;
    for (int v = from, d = 0;;) {
      int u = v;
      for (int to : gr[v]) {
        if (to != prnt[v] && cnt[to] > mid) {
          v = to;
          break;
        }
      }
      if (v == u) {
        if (v == from) {
          from = AskNext(from);
          --dist;
          break;
        }
        int nd = AskDist(v);
        if (nd == dist - d) {
          from = v;
          dist = nd;
          break;
        }
        int ffd = (nd - dist + d) >> 1;
        if (ffd == d) {
          from = AskNext(from);
          --dist;
          break;
        }
        for (int i = 0; i < (ffd); ++i) {
          v = prnt[v];
        }
        from = v;
        dist += ffd - d;
        break;
      }
      ++d;
    }
  }
  return 0;
}