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;
vector<int> g[200005];
bool mk[200005], mk2[200005], ok[200005];
int p[200005], l[200005], sons[200005][2];
void dfs(int u) {
  mk[u] = 1;
  for (auto y : g[u])
    if (!mk[y] && !mk2[y]) {
      l[y] = l[u] + 1;
      dfs(y);
      p[y] = u;
    }
}
void dfs1(int u) {
  mk[u] = 1;
  if (ok[u]) sons[u][0]++;
  for (auto y : g[u])
    if (!mk[y] && !mk2[y]) {
      dfs1(y);
      sons[u][0] += sons[y][0];
      sons[u][1] = max(sons[y][0], sons[u][1]);
    }
}
int main() {
  int n, u, v;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  p[0] = -1;
  dfs(0);
  cout << "d 1" << '\n';
  cout.flush();
  int level;
  cin >> level;
  int cont = 0;
  for (int i = 0; i < n; i++)
    if (l[i] == level) {
      ok[i] = 1;
      cont++;
    }
  for (int i = 0; i < n; i++) mk[i] = 0;
  int r = 0;
  while (cont > 1) {
    dfs1(r);
    for (int i = 0; i < n; i++)
      if (sons[i][0] > cont / 2 && sons[i][1] <= cont / 2) {
        cout << "d" << ' ' << i + 1 << '\n';
        cout.flush();
        int x;
        cin >> x;
        if (x == level - l[i]) {
          mk2[i] = 1;
          cout << "s" << ' ' << i + 1 << '\n';
          cout.flush();
          cin >> r;
          r--;
          cont = sons[r][0];
        } else {
          mk2[i] = 1;
          cont -= sons[i][0];
        }
        break;
      }
    for (int i = 0; i < n; i++) mk[i] = sons[i][0] = sons[i][1] = 0;
  }
  dfs1(r);
  for (int i = 0; i < n; i++)
    if (mk[i] && ok[i]) {
      cout << "! " << i + 1 << '\n';
      break;
    }
  return 0;
}