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;
vector<int> *g;
struct cc {
  int w, u, h;
};
const struct cc cendef = {1000000, -1, -1};
struct cc cen = cendef;
struct cc min(struct cc a, struct cc b) {
  if (a.w < b.w)
    return a;
  else
    return b;
}
int dist, root;
vector<int> si;
vector<bool> block;
int find_centre(int u, int s, int hi = 0, int p = -1) {
  int count = 1, m = 0;
  for (auto i : g[u]) {
    if (i == p || block[i]) continue;
    int k = find_centre(i, s, hi + 1, u);
    count += k;
    m = max(m, k);
  }
  m = max(m, s - count);
  struct cc r;
  r.w = m;
  r.u = u;
  r.h = hi;
  cen = min(cen, r);
  si[u] = count;
  return count;
}
void decomp(int s) {
  if (dist == 0) return;
  cen = cendef;
  find_centre(root, s);
  int c = cen.u;
  int dc = cen.h, cx;
  cout << "d " << c << endl;
  cout << flush;
  cin >> cx;
  if (cx == 0) {
    root = c;
    return;
  }
  block[c] = true;
  if (cx == dist - dc) {
    cout << "s " << c << endl;
    cout << flush;
    cin >> root;
    dist = cx - 1;
    decomp(si[root]);
  } else {
    decomp(s - si[c]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, i, j, k;
  cin >> n;
  g = new vector<int>[n + 1];
  block.assign(n + 1, false);
  si.resize(n + 1);
  for (i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cout << "d " << 1 << endl;
  cout << flush;
  cin >> dist;
  root = 1;
  decomp(n);
  cout << "! " << root << endl;
  cout << flush;
}