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 long long int N = 2e5 + 5;
long long int sub[N], mxid[N], lev[N], par[N];
vector<long long int> G[N];
long long int dfs(long long int a, long long int p) {
  par[a] = p;
  sub[a] = 1;
  long long int mx = 0, id = -1;
  for (long long int i = 0; i < G[a].size(); ++i) {
    long long int x = G[a][i];
    if (x != p) {
      lev[x] = lev[a] + 1;
      long long int r = dfs(x, a);
      if (r > mx) {
        mx = r;
        id = x;
      }
      sub[a] += r;
    }
  }
  mxid[a] = id;
  return sub[a];
}
long long int d;
long long int solve(long long int a) {
  if (lev[a] == d) {
    return a;
  }
  long long int cur = a;
  long long int dep = lev[cur];
  long long int rch = dep;
  for (long long int i = dep + 1; i < d + 1; ++i) {
    if (mxid[cur] != -1) {
      cur = mxid[cur];
      rch = i;
    } else {
      break;
    }
  }
  cout << "d " << cur << endl;
  long long int ab;
  cin >> ab;
  if (ab == 0) {
    return cur;
  }
  for (long long int i = rch - 1; i >= dep; i--) {
    cur = par[cur];
    if (d - lev[cur] + rch - i == ab) {
      cout << "s " << cur << endl;
      cin >> cur;
      return solve(cur);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i < n - 1; ++i) {
    long long int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  lev[1] = 0;
  dfs(1, 0);
  cout << "d " << 1 << endl;
  long long int a;
  cin >> a;
  if (a == 0) {
    cout << "! " << 1 << endl;
    return 0;
  }
  d = a;
  cout << "! " << solve(1) << endl;
  return 0;
}