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 inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int mod = 1000000007;
int modpow(int x, int y, int md = mod) {
  if (y == 0) return 1;
  int ret = modpow(x, y >> 1, md);
  ret = (long long)ret * ret % md;
  if (y & 1) ret = (long long)ret * x % md;
  return ret;
}
int n, cur = 1, deep;
vector<int> g[200005];
int dep[200005], sum[200005], Max[200005], p[200005], chain[200005];
void dfs(int x, int par = 0, int depth = 0) {
  p[x] = par;
  dep[x] = depth;
  sum[x] = (dep[x] == deep ? 1 : 0);
  Max[x] = -inf;
  chain[x] = x;
  for (auto to : g[x])
    if (to != par) {
      dfs(to, x, depth + 1);
      sum[x] += sum[to];
      if (sum[to] > Max[x]) {
        Max[x] = sum[to];
        chain[x] = chain[to];
      }
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  cout << "d " << 1 << endl;
  fflush(stdout);
  cin >> deep;
  if (deep == 0) {
    cout << "! 1" << endl;
    fflush(stdout);
    return 0;
  }
  dfs(1);
  int limit = 50;
  while (limit--) {
    int tmp = chain[cur], len = dep[tmp];
    cout << "d " << tmp << endl;
    fflush(stdout);
    int v;
    cin >> v;
    int lca = (len + deep - v) >> 1;
    for (int i = dep[cur]; i < lca; i++) {
      for (auto to : g[cur]) {
        if (to != p[cur] && chain[cur] == chain[to]) {
          cur = to;
          break;
        }
      }
    }
    if (dep[cur] == deep) {
      cout << "! " << cur << endl;
      fflush(stdout);
      return 0;
    }
    cout << "s " << cur << endl;
    fflush(stdout);
    cin >> cur;
  }
  return 0;
}