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 void _read(T& x) {
  cin >> x;
}
void R() {}
template <typename T, typename... U>
void R(T& head, U&... tail) {
  _read(head);
  R(tail...);
}
template <typename T>
inline void _write(const T& x) {
  cout << x << ' ';
}
void W() { cout << endl; }
template <typename T, typename... U>
void W(const T& head, const U&... tail) {
  _write(head);
  W(tail...);
}
template <typename T>
void debug(vector<T> V) {
  for (auto x : V) {
    cout << x << " ";
  }
  cout << endl;
}
void go();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  go();
  return 0;
}
const int MAXN = 2e5 + 10;
vector<int> G[MAXN];
int dep[MAXN], sz[MAXN];
int depx;
int ask(char op, int u) {
  cout << op << " " << u << endl;
  int res;
  R(res);
  return res;
}
int pre(int u, int f) {
  sz[u] = 1;
  for (auto v : G[u]) {
    if (v == f) {
      continue;
    }
    dep[v] = dep[u] + 1;
    sz[u] += pre(v, u);
  }
  return sz[u];
}
vector<int> h;
void get(int u) {
  h.push_back(u);
  int big = -1;
  for (auto v : G[u]) {
    if (dep[v] > dep[u] && (big == -1 || sz[v] > sz[big])) {
      big = v;
    }
  }
  if (big != -1) {
    get(big);
  }
}
int dfs(int u) {
  h.clear();
  get(u);
  int depy = (depx + dep[h.back()] - ask('d', h.back())) / 2;
  int v = h[depy - dep[u]];
  if (depx == depy) {
    return v;
  }
  return dfs(ask('s', v));
}
void go() {
  int n;
  R(n);
  for (int i = 2; i <= n; i++) {
    int x, y;
    R(x, y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  depx = ask('d', 1);
  pre(1, -1);
  cout << "! " << dfs(1) << endl;
}