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;
int n;
int last[400007], other[400007], pre[400007], cnt = 0;
int max_son[400007], d[400007], size[400007], dep[400007];
int now;
void insert(int u, int v) {
  other[++cnt] = v, pre[cnt] = last[u], last[u] = cnt;
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  size[u] = 1;
  d[u] = u;
  for (int q = last[u]; q; q = pre[q]) {
    int v = other[q];
    if (v != fa) {
      dfs(v, u);
      size[u] += size[v];
      if (size[v] > size[max_son[u]]) {
        max_son[u] = v;
        d[u] = d[v];
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    insert(u, v), insert(v, u);
  }
  dep[0] = -1;
  dfs(1, 0);
  printf("d %d\n", 1);
  fflush(stdout);
  scanf("%d", &now);
  int u = 1;
  while (dep[u] != now) {
    printf("d %d\n", d[u]);
    fflush(stdout);
    int dis;
    scanf("%d", &dis);
    dis = (now + dep[d[u]] - dis) / 2 - dep[u];
    for (int i = 1; i <= dis; i++) u = max_son[u];
    if (dep[u] != now) {
      printf("s %d\n", u);
      fflush(stdout);
      scanf("%d", &u);
    }
  }
  printf("! %d\n", u);
  fflush(stdout);
  return 0;
}