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;
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
long long modExpo(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modExpo((x * x) % 1000000007, n / 2);
  else
    return (x * modExpo((x * x) % 1000000007, (n - 1) / 2)) % 1000000007;
}
long long modInverse(long long n) { return modExpo(n, 1000000007 - 2); }
long long n, m, k, ar[200010], dx, y, dy, dxy;
string s, s1;
vector<long long> adj[200010];
long long sz[200010], max_size_child[200010], dep[200010], la[200010];
void dfs(long long ver, long long par, long long l) {
  dep[ver] = l;
  long long chi = ver, chi_sz = 0;
  for (auto i : adj[ver]) {
    if (i != par) {
      dfs(i, ver, l + 1);
      sz[ver] += (sz[i] + 1);
      if (sz[i] + 1 > chi_sz) {
        chi = i;
        chi_sz = sz[i] + 1;
      }
    }
  }
  if (chi_sz == 0)
    la[ver] = chi;
  else
    la[ver] = la[chi];
  max_size_child[ver] = chi;
}
long long find(long long ver, long long l) {
  while (dep[ver] != l) ver = max_size_child[ver];
  return ver;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int ti = 1;
  while (ti--) {
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    dfs(1, 1, 0);
    long long u = 1, temp;
    cout << "d " << u << endl << flush;
    cin >> dx;
    {
      if (dx == -1) return 0;
    }
    while (1) {
      if (dx == 0) {
        cout << "! " << u << endl << flush;
        return 0;
      }
      y = la[u];
      cout << "d " << y << endl << flush;
      cin >> dxy;
      {
        if (dxy == -1) return 0;
      }
      if (dxy == 0) {
        cout << "! " << y << endl << flush;
        return 0;
      }
      dy = dep[y] - dep[u];
      k = (dx + dy - dxy) / 2;
      temp = find(u, k + dep[u]);
      if (k == dx) {
        cout << "! " << temp << endl << flush;
        return 0;
      }
      cout << "s " << temp << endl << flush;
      cin >> u;
      {
        if (u == -1) return 0;
      }
      dx = dx - k - 1;
    }
  }
  return 0;
}