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;
using ll = long long;
using vll = vector<ll>;
using vii = vector<int>;
using pii = pair<int, int>;
template <typename T = vii>
using vec = vector<T>;
const ll INF = 1e18, MOD = 1e9 + 7;
const int MAX = 1e6;
vec<> adj;
vii depth, s_size;
int dfs0(int curr = 0, int prev = -1, int d = 0) {
  depth[curr] = d;
  s_size[curr] = 1;
  for (int node : adj[curr]) {
    if (node == prev) {
      continue;
    }
    s_size[curr] += dfs0(node, curr, d + 1);
  }
  return s_size[curr];
}
vii next_node, last;
int dfs1(int curr = 0, int prev = -1) {
  int max_sub = 0;
  next_node[curr] = -1, last[curr] = curr;
  for (int node : adj[curr]) {
    if (node == prev) {
      continue;
    }
    if (s_size[node] > max_sub) {
      next_node[curr] = node;
      last[curr] = dfs1(node, curr);
      max_sub = s_size[node];
    } else {
      dfs1(node, curr);
    }
  }
  return last[curr];
}
int dist(int u, int v) { return abs(depth[u] - depth[v]); }
int dist(int u) {
  cout << "d " << (u + 1) << endl;
  int res;
  cin >> res;
  return res;
}
int get_node(int u, int d) {
  while (d) u = next_node[u], d--;
  return u;
}
int get_node(int u) {
  cout << "s " << (u + 1) << endl;
  int res;
  cin >> res;
  return res - 1;
}
void output(int u) { cout << "! " << (u + 1) << endl; }
int main() {
  int n;
  cin >> n;
  adj.resize(n);
  depth.resize(n);
  s_size.resize(n);
  next_node.resize(n);
  last.resize(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);
  }
  dfs0();
  dfs1();
  int u = 0, x = dist(u);
  while (true) {
    if (x == 0) {
      output(u);
      break;
    }
    const int v = last[u], y = dist(v), z = dist(u, v), b = (x + y - z) / 2;
    if (b == 0) {
      output(get_node(u, z - y));
      break;
    }
    int w = get_node(u, x - b);
    u = get_node(w), x = b - 1;
  }
}