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;
const int maxn = 2e5 + 10;
int n, u, vv, root = 1, dis[maxn], disx, siz[maxn], x1;
vector<int> v[maxn], path, save;
vector<pair<int, int> > ord;
queue<int> q;
void bfs(int node) {
  for (int i = 0; i < v[node].size(); i++) {
    if (dis[node] + 1 < dis[v[node][i]]) {
      q.push(v[node][i]);
      dis[v[node][i]] = dis[node] + 1;
    }
  }
  q.pop();
  if (q.empty()) return;
  bfs(q.front());
}
void bigpath(int node) {
  path.push_back(node);
  int child = 0;
  for (int i = 0; i < v[node].size(); i++) {
    if (dis[node] < dis[v[node][i]]) {
      if (child == 0)
        child = v[node][i];
      else if (siz[child] < siz[v[node][i]])
        child = v[node][i];
    }
  }
  if (child == 0) return;
  bigpath(child);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> vv;
    v[vv].push_back(u);
    v[u].push_back(vv);
  }
  for (int i = 1; i <= n; i++) dis[i] = maxn;
  dis[1] = 0;
  q.push(1);
  bfs(1);
  for (int i = 1; i <= n; i++) ord.push_back({dis[i], i});
  sort(ord.begin(), ord.end());
  for (int i = ord.size() - 1; i >= 0; i--) {
    siz[ord[i].second] = 1;
    for (int i2 = 0; i2 < v[ord[i].second].size(); i2++) {
      if (ord[i].first < dis[v[ord[i].second][i2]])
        siz[ord[i].second] += siz[v[ord[i].second][i2]];
    }
  }
  cout << "d 1" << endl;
  cout.flush();
  cin >> disx;
  while (1) {
    path.clear();
    bigpath(root);
    cout << "d " << path[path.size() - 1] << endl;
    cout.flush();
    cin >> x1;
    root = path[(dis[path[path.size() - 1]] + disx - x1) / 2];
    if (dis[root] == disx) {
      cout << "! " << root << endl;
      cout.flush();
      return 0;
    }
    cout << "s " << root << endl;
    cout.flush();
    cin >> root;
    int w = dis[root];
    disx -= w;
    for (int i = 1; i <= n; i++) dis[i] -= w;
  }
  return 0;
}