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 MAXN = 200005;
int n;
vector<int> par(MAXN, -1);
vector<int> pos(MAXN, -1);
vector<vector<int> > g;
vector<vector<int> > h;
vector<vector<int> > tb;
int d = 0;
void dfs(int u, int p, int tall) {
  if (h.size() <= tall) h.push_back(vector<int>());
  h[tall].push_back(u);
  par[u] = p;
  pos[u] = h[tall].size() - 1;
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u, tall + 1);
    }
  }
}
void initTable() {
  tb.assign(n, vector<int>(20, -1));
  for (int i = 0; i < n; ++i) tb[i][0] = par[i];
  for (int i = 1; i < 20; ++i) {
    for (int j = 0; j < n; ++j) {
      if (tb[j][i - 1] != -1) tb[j][i] = tb[tb[j][i - 1]][i - 1];
    }
  }
}
int lift(int u, int k) {
  for (int i = 19; i >= 0; --i) {
    if (k >= (1 << i)) {
      k -= (1 << i);
      u = tb[u][i];
    }
  }
  return u;
}
signed main() {
  cin >> n;
  g.assign(n, vector<int>());
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1, 0);
  initTable();
  cout << "d 1" << endl;
  cin >> d;
  int l = 0, r = h[d].size() - 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    cout << "d " << h[d][mid] + 1 << endl;
    int dist;
    cin >> dist;
    if (dist == 0) {
      cout << "! " << h[d][mid] + 1 << endl;
      return 0;
    }
    dist /= 2;
    int u = lift(h[d][mid], dist);
    int v = lift(h[d][mid], dist - 1);
    cout << "s " << u + 1 << endl;
    int uu;
    cin >> uu;
    --uu;
    if (pos[v] > pos[uu]) {
      r = mid - 1;
    } else
      l = mid + 1;
  }
}