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> heavy, parent, head, size_, end_, pos, orgpos;
vector<list<int> > adj;
int curr = 1;
void dfs(int i) {
  size_[i] = 1;
  int maxsize_ = 0;
  for (auto it : adj[i]) {
    if (it != parent[i]) {
      parent[it] = i;
      dfs(it);
      size_[i] += size_[it];
      if (size_[it] > maxsize_) {
        maxsize_ = size_[it];
        heavy[i] = it;
      }
    }
  }
}
void decompose(int i, int h) {
  head[i] = h;
  pos[i] = curr;
  orgpos[curr] = i;
  curr++;
  if (heavy[i] != -1) {
    decompose(heavy[i], h);
  } else {
    end_[h] = i;
  }
  for (auto it : adj[i]) {
    if (it != heavy[i] && it != parent[i]) decompose(it, it);
  }
}
int32_t main() {
  int n, u, v;
  cin >> n;
  heavy = vector<int>(n + 1, -1);
  pos = vector<int>(n + 1);
  orgpos = vector<int>(n + 1);
  parent = vector<int>(n + 1, 0);
  head = vector<int>(n + 1);
  end_ = vector<int>(n + 1);
  size_ = vector<int>(n + 1);
  adj = vector<list<int> >(n + 1);
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  decompose(1, 1);
  int k1 = 1, k2, x, y, d, nd = -1, z, k3;
  printf("d %d\n", k1);
  fflush(stdout);
  scanf("%d", &x);
  while (1) {
    x = x - nd - 1;
    k2 = end_[k1];
    if (k2 != k1) {
      printf("d %d\n", k2);
      fflush(stdout);
      scanf("%d", &y);
    } else
      y = x;
    if (y == -1) exit(0);
    d = pos[k2] - pos[k1];
    nd = (x + d - y) / 2;
    k1 = orgpos[pos[k1] + nd];
    if (nd == x) {
      printf("! %d\n", k1);
      fflush(stdout);
      break;
    }
    printf("s %d\n", k1);
    fflush(stdout);
    scanf("%d", &z);
    if (z == -1) exit(0);
    k1 = z;
  }
  return 0;
}