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;
set<int> AdjList[200001];
int nChild[200001];
int d[200001];
int p[200001];
int d_x;
int cur_sz = 0;
int lim;
void cal_d(int u, int par) {
  for (set<int>::iterator it = AdjList[u].begin(); it != AdjList[u].end();
       it++) {
    int v = *it;
    if (v != par) {
      p[v] = u;
      d[v] = d[u] + 1;
      cal_d(v, u);
    }
  }
}
void cal_sz(int u, int par) {
  cur_sz++;
  nChild[u] = 1;
  for (set<int>::iterator it = AdjList[u].begin(); it != AdjList[u].end();
       it++) {
    int v = *it;
    if (v != par) {
      cal_sz(v, u);
      nChild[u] += nChild[v];
    }
  }
}
int find_c(int u, int par) {
  for (set<int>::iterator it = AdjList[u].begin(); it != AdjList[u].end();
       it++) {
    int v = *it;
    if (v != par && nChild[v] > lim) return find_c(v, u);
  }
  return u;
}
void decomp(int u) {
  cur_sz = 0;
  cal_sz(u, -1);
  lim = cur_sz / 2;
  int c = find_c(u, -1);
  int cur_d;
  cout << "d " << c << endl;
  cin >> cur_d;
  if (cur_d == 0) {
    cout << "! " << c << endl;
    exit(0);
  }
  if (d_x - d[c] == cur_d) {
    cout << "s " << c << endl;
    int nxt;
    cin >> nxt;
    AdjList[nxt].erase(c);
    decomp(nxt);
  } else {
    AdjList[p[c]].erase(c);
    decomp(p[c]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    AdjList[u].insert(v);
    AdjList[v].insert(u);
  }
  cal_d(1, -1);
  cout << "d 1" << endl;
  cin >> d_x;
  decomp(1);
}