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>
#pragma GCC optimize(3)
using namespace std;
const int N = 2e5 + 5;
vector<int> G[N], G2[N];
int n, m, u, v;
int answer(int u) {
  printf("! %d\n", u);
  fflush(stdout);
  return 0;
}
int Q1(int u) {
  printf("d %d\n", u);
  fflush(stdout);
  scanf("%d", &u);
  if (u == -1) exit(0);
  return u;
}
int Q2(int u) {
  printf("s %d\n", u);
  fflush(stdout);
  scanf("%d", &u);
  if (u == -1) exit(0);
  return u;
}
int vis[N];
vector<int> V;
int dfs(int u, int d) {
  if (!d) {
    V.push_back(u);
    return 1;
  }
  int rr = 0;
  for (auto &v : G[u]) {
    if (vis[v]) continue;
    vis[v] = u;
    if (dfs(v, d - 1)) {
      G2[u].push_back(v);
      rr = 1;
    }
  }
  return rr;
}
void re_dfs(int u, int d) {
  memset(vis, 0, sizeof(vis));
  vis[u] = -1;
  V.clear();
  for (int i = 1; i <= n; ++i) G2[i].clear();
  dfs(u, d);
  swap(G2, G);
}
int main() {
  srand(time(0) + 233);
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  int dis = Q1(1), i = 1;
  re_dfs(1, dis);
  while (G[i].size() == 1) i = G[i][0], --dis;
  while (V.size() > 1) {
    i = Q2(i);
    --dis;
    re_dfs(i, dis);
    if (V.size() == 1) break;
    i = V[rand() % V.size()];
    dis = Q1(i) >> 1;
    for (int j = 0; j < dis; ++j) i = vis[i];
    re_dfs(i, dis);
  }
  return answer(V[0]);
}