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> vec[200001], my, tmp;
int visited[200001], dist[200001];
void dd(int cur, int cnt, int p) {
  if (cnt == 0) {
    visited[cur]++;
    return;
  }
  for (int v : vec[cur]) {
    if (p != v) dd(v, cnt - 1, cur);
  }
}
int find(int cur, int &nn) {
  int ret = visited[cur] == 2;
  for (int v : vec[cur]) {
    if (dist[v] > dist[cur]) {
      ret += find(v, nn);
    }
  }
  if (ret * 2 >= (int)my.size()) {
    nn = cur;
    return -1000;
  }
  return ret;
}
void init(int cur, int cnt, int p) {
  dist[cur] = cnt;
  for (int v : vec[cur]) {
    if (p != v) init(v, cnt + 1, cur);
  }
}
int main() {
  int n, a, b, now = 1, next, nwd, nxd;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    vec[a].push_back(b);
    vec[b].push_back(a);
    my.push_back(i);
  }
  init(1, 0, 1);
  my.push_back(n);
  bool flag = true;
  while (true) {
    if (flag) {
      printf("d %d\n", now);
      fflush(stdout);
      scanf("%d", &nwd);
      memset(visited, 0, sizeof(visited));
      dd(now, nwd, now);
      tmp.clear();
      for (int v : my) {
        if (visited[v]) {
          tmp.push_back(v);
          visited[v]++;
        }
      }
      my = tmp;
      flag = false;
    }
    if (my.size() == 1) {
      printf("! %d", my[0]);
      fflush(stdout);
      return 0;
    }
    find(now, next);
    if (now == next) {
      printf("s %d\n", now);
      fflush(stdout);
      scanf("%d", &now);
      flag = true;
      continue;
    }
    printf("d %d\n", next);
    fflush(stdout);
    scanf("%d", &nxd);
    if (dist[next] - dist[now] == nwd - nxd) {
      now = next;
      nwd = nxd;
    }
    memset(visited, 0, sizeof(visited));
    dd(next, nxd, next);
    tmp.clear();
    for (int v : my) {
      if (visited[v]) {
        tmp.push_back(v);
        visited[v]++;
      }
    }
    my = tmp;
  }
  return 0;
}