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.
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g(200005);
int d[200005];
int p[200005];
int Size[200005];
bool del[200005];
int depth_x;
int query(string s, int node) {
  cout << s << " " << node << "\n";
  cout.flush();
  int res;
  cin >> res;
  if (res == -1) {
    exit(0);
  }
  return res;
}
void dfs(int node, int par, int depth) {
  d[node] = depth;
  p[node] = par;
  for (auto it : g[node]) {
    if (it == par) {
      continue;
    }
    dfs(it, node, depth + 1);
  }
}
int getSize(int node, int par) {
  Size[node] = 1;
  for (auto it : g[node]) {
    if (it == par || del[it]) {
      continue;
    }
    Size[node] += getSize(it, node);
  }
  return Size[node];
}
int find_c(int node, int par, int sz) {
  for (auto it : g[node]) {
    if (it == par || del[it]) {
      continue;
    }
    if (Size[it] > (sz / 2)) {
      return find_c(it, node, sz);
    }
  }
  return node;
}
void solve(int node) {
  int sz = getSize(node, 0);
  int centroid = find_c(node, 0, sz);
  del[centroid] = 1;
  int dist = query("d", centroid);
  if (dist == 0) {
    cout << "! " << centroid << "\n";
    cout.flush();
    exit(0);
  }
  if (depth_x - d[centroid] == dist) {
    solve(query("s", centroid));
  } else {
    solve(p[centroid]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  depth_x = query("d", 1);
  dfs(1, 0, 0);
  solve(1);
}
