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;
const int N = 2e5 + 5;
int n, h[N], sz[N], par[N];
int dist[N], nxt[N];
vector<int> adj[N];
bool mark[N];
void dfs(int u) {
  sz[u] = 1;
  for (auto v : adj[u])
    if (!mark[v] && v ^ par[u]) {
      h[v] = h[par[v] = u] + 1;
      dfs(v), sz[u] += sz[v];
    }
}
inline int get_center(int root) {
  dfs(root);
  for (int u = root, big = -1; true; u = big, big = -1) {
    for (auto v : adj[u])
      if (!mark[v] && v ^ par[u] && (!~big || sz[v] > sz[big])) big = v;
    if (max(~big ? sz[big] : 0, sz[root] - sz[u]) <= sz[root] >> 1) return u;
  }
}
inline int get_dist(int u) {
  if (dist[u]) return dist[u];
  cout << "d " << u + 1 << endl;
  cin >> dist[u];
  if (!dist[u]) {
    cout << "! " << ++u;
    exit(0);
  }
  return dist[u];
}
inline int get_nxt(int u) {
  if (nxt[u]) return nxt[u];
  cout << "s " << u + 1 << endl;
  cin >> nxt[u];
  return --nxt[u];
}
inline void read_input() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
inline void solve(int u = 0) {
  int v = get_center(u);
  if (get_dist(v) ^ h[u] + get_dist(u) - h[v]) {
    mark[v] = true;
    solve(u);
  }
  if (u ^ v) {
    mark[par[v]] = true;
    solve(v);
  }
  mark[u] = true;
  solve(get_nxt(u));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), solve();
  return 0;
}