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;
vector<int> graf[2000010];
int gle[2000010];
int P[2000010];
int pod[2000010];
int odl_1;
void dfs(int v) {
  pod[v] = 1;
  for (auto x : graf[v]) {
    if (pod[x] == 0) {
      P[x] = v;
      gle[x] = gle[v] + 1;
      dfs(x);
      pod[v] += pod[x];
    }
  }
}
void centr(int v) {
  for (auto x : graf[v]) {
    if (pod[x] * 2 > pod[v]) {
      P[v] = x;
      P[x] = x;
      int cat = pod[v];
      pod[v] -= pod[x];
      pod[x] = cat;
      centr(x);
      return;
    }
  }
  cout << "d " << v << endl;
  cout.flush();
  int par;
  cin >> par;
  if (par == 0) {
    cout << "! " << v << endl;
    cout.flush();
    exit(0);
  }
  if (par + gle[v] == odl_1) {
    cout << "s " << v << endl;
    cout.flush();
    int x;
    cin >> x;
    if (x == -1) exit(0);
    P[x] = x;
    pod[v] = 0;
    centr(x);
  } else {
    int szuk;
    for (auto x : graf[v]) {
      if (gle[x] < gle[v]) szuk = x;
    }
    P[szuk] = szuk;
    pod[v] = 0;
    centr(szuk);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < (n - 1); ++i) {
    int t1, t2;
    cin >> t1 >> t2;
    graf[t1].push_back(t2);
    graf[t2].push_back(t1);
  }
  cout << "d 1" << endl;
  cout.flush();
  cin >> odl_1;
  if (odl_1 == 0) {
    cout << "! 1" << endl;
    cout.flush();
    return 0;
  }
  P[1] = 1;
  dfs(1);
  centr(1);
  return 0;
}