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 bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
inline void read(T &x) {
  char ch = getchar();
  int f = 1;
  x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x), read(args...);
}
int n, dep[200050], fa[200050], size[200050], ms[200050], rt, Size, vis[200050];
int Dep;
vector<int> adj[200050];
void getd(int u) {
  for (int v : adj[u])
    if (v != fa[u]) fa[v] = u, dep[v] = dep[u] + 1, getd(v);
}
void getrt(int u, int f) {
  size[u] = 1, ms[u] = 0;
  for (int v : adj[u])
    if (v != f && !vis[v])
      getrt(v, u), chkmax(ms[u], size[v]), size[u] += size[v];
  chkmax(ms[u], Size - size[u]);
  if (ms[u] < ms[rt]) rt = u;
}
int askd(int u) {
  int x;
  cout << "d " << u << endl, fflush(stdout);
  cin >> x;
  return x;
}
int asku(int u) {
  int x;
  cout << "s " << u << endl, fflush(stdout);
  cin >> x;
  return x;
}
void Answer(int u) {
  cout << "! " << u << endl, fflush(stdout);
  exit(0);
}
void solve(int u) {
  getrt(u, 0), vis[u] = 1;
  int _d1 = askd(u), v;
  if (_d1 == 0) Answer(u);
  if (_d1 + dep[u] == Dep)
    v = asku(u);
  else
    v = fa[u];
  Size = size[v], rt = 0, getrt(v, 0), solve(rt);
}
int main() {
  read(n);
  int u, v;
  for (register int i = 1; i <= n - 1; ++i)
    read(u, v), adj[u].push_back(v), adj[v].push_back(u);
  getd(1), Dep = askd(1);
  Size = n, ms[0] = 1e9, rt = 0, getrt(1, 0), solve(rt);
  return 0;
}