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 dist1;
vector<int> G[200009];
int si[200009];
int heavy[200009];
bool was[200009];
int lev[200009];
void dfs(int x) {
  was[x] = true;
  si[x]++;
  for (int sz : G[x]) {
    if (!was[sz]) {
      lev[sz] = lev[x] + 1;
      dfs(sz);
      si[x] += si[sz];
      if (heavy[x] == 0 || si[sz] > si[heavy[x]]) heavy[x] = sz;
    }
  }
}
void sch(int x) {
  if (si[x] == 1 || lev[x] == dist1) {
    cout << "! " << x << endl;
    fflush(stdout);
    exit(0);
  }
  int leaf = x;
  while (si[leaf] > 1) leaf = heavy[leaf];
  cout << "d " << leaf << endl;
  fflush(stdout);
  int dl;
  cin >> dl;
  if (dl == 0) {
    cout << "! " << leaf << endl;
    fflush(stdout);
    exit(0);
  }
  int lcalev = (dist1 + lev[leaf] - dl) / 2;
  while (lev[x] + 1 < lcalev) x = heavy[x];
  cout << "s " << x << endl;
  fflush(stdout);
  int direction;
  cin >> direction;
  sch(direction);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1);
  cout << "d 1" << endl;
  fflush(stdout);
  cin >> dist1;
  if (dist1 == 0) {
    cout << "! 1" << endl;
    fflush(stdout);
    return 0;
  }
  sch(1);
  return 0;
}