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 = 5E5 + 5;
vector<int> graph[N];
int n, dep[N], sz[N], par[N], plca[N][25];
void dfs(int src, int p, int h) {
  par[src] = p;
  dep[src] = h;
  sz[src] = 1;
  for (auto k : graph[src]) {
    if (k == p) continue;
    dfs(k, src, h + 1);
    sz[src] += sz[k];
  }
}
int get_hld(int src) {
  int mx = 0, child = src;
  for (auto k : graph[src]) {
    if (k == par[src]) continue;
    if (sz[k] > mx) {
      mx = sz[k];
      child = k;
    }
  }
  if (mx != 0) {
    child = get_hld(child);
  }
  return child;
}
void buildlca() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 25; j++) plca[i][j] = -1;
  for (int i = 0; i < n; i++) plca[i][0] = par[i];
  for (int j = 1; j < 25; j++)
    for (int i = 0; i < n; i++)
      if (plca[i][j - 1] != -1) plca[i][j] = plca[plca[i][j - 1]][j - 1];
}
int lca(int v, int depy, int depu) {
  depy += depu;
  if (dep[v] == depy) return v;
  for (int j = 24; j >= 0; j--) {
    if (plca[v][j] != -1 && dep[plca[v][j]] >= depy) {
      v = plca[v][j];
    }
    if (dep[v] == depy) return v;
  }
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(0, -1, 0);
  buildlca();
  int u = 0;
  cout << "d " << u + 1 << endl << flush;
  int depx;
  cin >> depx;
  if (depx == -1) return 0;
  if (depx == 0) {
    cout << "! " << u + 1 << endl << flush;
    return 0;
  }
  int prev_depx = depx;
  int t = 18;
  while (t--) {
    depx = prev_depx - dep[u];
    if (depx == 0) {
      cout << "! " << u + 1 << endl << flush;
      return 0;
    }
    int v = get_hld(u);
    int depv = dep[v] - dep[u];
    cout << "d " << v + 1 << endl << flush;
    int distv;
    cin >> distv;
    if (distv == -1) return 0;
    if (distv == 0) {
      cout << "! " << v + 1 << endl << flush;
      return 0;
    }
    int depy = (depx + depv - distv) / 2;
    int y = lca(v, depy, dep[u]);
    if (depx == depy) {
      cout << "! " << y + 1 << endl << flush;
      return 0;
    }
    cout << "s " << y + 1 << endl << flush;
    int prevu = u;
    cin >> u;
    if (u == -1) return 0;
    u--;
    if (dep[u] == depx + dep[prevu]) {
      cout << "! " << u + 1 << endl << flush;
      return 0;
    }
  }
  return 0;
}