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;
vector<int> adj[200001];
int sss[200001];
int n;
int it[200001];
void dfs(int no, int par = -1) {
  sss[no] = it[no];
  for (auto j : adj[no]) {
    if (j != par) {
      dfs(j, no);
      sss[no] += sss[j];
    }
  }
}
int find(int no, int par = -1) {
  for (auto j : adj[no]) {
    if (sss[j] > sss[no] / 2 and j != par) {
      return find(j, no);
    }
  }
  return no;
}
vector<int> ss;
set<int> cur;
int cot = 0;
void dfs4(int no, int par = -1, int levv = 0) {
  if (levv == cot and it[no] == 1) {
    ss.push_back(no);
  }
  for (auto j : adj[no]) {
    if (j != par) {
      dfs4(j, no, levv + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int aa, bb;
    cin >> aa >> bb;
    adj[aa].push_back(bb);
    adj[bb].push_back(aa);
  }
  for (int i = 1; i <= n; i++) {
    it[i] = 1;
  }
  cout << "d " << 1 << endl;
  cin >> cot;
  if (cot == 0) {
    cout << "! " << 1 << endl;
    return 0;
  }
  dfs4(1);
  for (auto j : ss) {
    cur.insert(j);
  }
  int co = 1;
  while (cur.size() > 1) {
    for (int i = 1; i <= n; i++) {
      it[i] = 0;
    }
    for (auto j : cur) {
      it[j] = 1;
    }
    dfs(1);
    int cen = find(1);
    cout << "d " << cen << endl;
    cin >> cot;
    if (cot == 0) {
      cout << "! " << cen << endl;
      return 0;
    }
    ss.clear();
    dfs4(cen);
    int xx = cur.size();
    cur.clear();
    for (auto j : ss) {
      cur.insert(j);
    }
    co += 1;
    if (cur.size() <= ((xx + 1) / 2)) {
      continue;
    }
    cout << "s " << cen << endl;
    co += 1;
    cot -= 1;
    int x;
    cin >> x;
    ss.clear();
    dfs4(x);
    cur.clear();
    for (auto j : ss) {
      cur.insert(j);
    }
    if (cur.size() > ((xx + 1) / 2)) {
      while (true) {
        continue;
      }
    }
  }
  cout << "! " << (*(cur.begin())) << endl;
  return 0;
}