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 N = 2e5 + 5;
vector<int> g[N];
vector<int> p(N, -1);
vector<int> d(N, -1);
vector<bool> blc(N);
int dx;
int q(char op, int u) {
  cout << op << ' ' << u << endl;
  int r;
  cin >> r;
  return r;
}
void pre(int u) {
  for (int v : g[u]) {
    if (!~d[v]) {
      d[v] = d[u] + 1;
      p[v] = u;
      pre(v);
    }
  }
}
vector<int> sz(N);
int _n;
void comp_size(int u, int pa) {
  sz[u] = 1;
  for (int v : g[u]) {
    if (!blc[v] && v != pa) {
      comp_size(v, u);
      sz[u] += sz[v];
    }
  }
}
int fcd(int u, int pa) {
  for (int v : g[u]) {
    if (!blc[v] && v != pa && sz[v] > _n / 2) return fcd(v, u);
  }
  return u;
}
int cd(int u) {
  comp_size(u, 0);
  _n = sz[u];
  return fcd(u, 0);
}
int qcd(int u) {
  int v = cd(u);
  int dis = q('d', v);
  if (!dis) return v;
  blc[v] = true;
  if (d[v] + dis == dx) return qcd(q('s', v));
  return qcd(p[v]);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  dx = q('d', 1);
  d[1] = 0;
  pre(1);
  int res = qcd(1);
  cout << "! " << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  cout << endl;
}