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;
struct Edge {
  int to, next;
};
Edge E[400005];
int head[200005], Ecou;
int N;
void init() {
  Ecou = 0;
  memset(head, -1, sizeof(head));
}
inline void addEdge(int u, int v) {
  E[Ecou].to = v;
  E[Ecou].next = head[u];
  head[u] = Ecou++;
}
int ccc = 0;
inline int cmd(char c, int v) {
  ++ccc;
  assert(ccc <= 36);
  printf("%c %d\n", c, v);
  fflush(stdout);
  int ret;
  scanf("%d", &ret);
  assert(ret != -1);
  return ret;
}
int vis[200005];
int fa[200005];
int ans, ansrem, sum;
int nnn[200005], dep[200005];
int dfs1(int u, int f) {
  int ret = 1;
  fa[u] = f;
  int maxn = -1, rem = -1;
  for (int i = head[u]; i != -1; i = E[i].next)
    if (E[i].to != f && vis[E[i].to] == 0) {
      dep[E[i].to] = dep[u] + 1;
      int tmp = dfs1(E[i].to, u);
      ret += tmp;
      if (tmp > maxn) {
        maxn = tmp;
        rem = E[i].to;
      }
    }
  nnn[u] = rem;
  return ret;
}
vector<int> path;
int findPath(int r) {
  path.clear();
  dep[r] = 0;
  dfs1(r, -1);
  for (int i = r; i != -1; i = nnn[i]) path.push_back(i);
  return *(path.rbegin());
}
int main() {
  init();
  scanf("%d", &N);
  for (int i = 1; i < N; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    addEdge(a, b);
    addEdge(b, a);
  }
  int depx = cmd('d', 1);
  int root = 1;
  while (1) {
    if (depx == 0) {
      printf("! %d\n", root);
      break;
    }
    int v = findPath(root);
    int dis1 = cmd('d', v);
    if (dis1 == 0) {
      printf("! %d\n", v);
      break;
    }
    int depy = (depx + dep[v] - dis1) / 2;
    int y = path[depy];
    assert(dep[y] == depy);
    if (depy == depx) {
      printf("! %d\n", y);
      break;
    }
    vis[y] = 1;
    root = cmd('s', y);
    depx -= depy + 1;
  }
  return 0;
}