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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> adj((size_t)n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<int> visit_vertexes, visit_depths, visit_enter((size_t)n),
      visit_leave((size_t)n), visit_sizes((size_t)n, 1),
      visit_next((size_t)n, -1);
  int current_time = 0, current_depth = 0;
  function<void(int, int)> visit = [&](int i, int j) {
    current_depth++;
    visit_vertexes.push_back(j);
    visit_depths.push_back(current_depth);
    visit_enter[j] = current_time++;
    bool first = true;
    for (int k : adj[j]) {
      if (k != i) {
        if (!first) {
          visit_vertexes.push_back(j);
          visit_depths.push_back(current_depth);
          current_time++;
        }
        first = false;
        visit(j, k);
        visit_sizes[j] += visit_sizes[k];
        if (visit_next[j] == -1 ||
            visit_sizes[k] > visit_sizes[visit_next[j]]) {
          visit_next[j] = k;
        }
      }
    }
    visit_vertexes.push_back(j);
    visit_depths.push_back(current_depth);
    visit_leave[j] = current_time++;
    current_depth--;
  };
  visit(-1, 0);
  int sparse_table_size = 0;
  while ((1 << sparse_table_size) <= current_time) {
    sparse_table_size++;
  }
  auto sparse_table_combine = [&](int j, int k) {
    if (j >= 0 && k >= 0) {
      return visit_depths[j] < visit_depths[k] ? j : k;
    } else {
      exit(-1);
    }
  };
  vector<vector<int>> sparse_table((size_t)sparse_table_size,
                                   vector<int>((size_t)current_time, -1));
  iota(sparse_table[0].begin(), sparse_table[0].end(), 0LL);
  for (int i = 1; i < sparse_table_size; i++) {
    for (int j = 0; j + (1 << i) <= current_time; j++) {
      sparse_table[i][j] = sparse_table_combine(
          sparse_table[i - 1][j], sparse_table[i - 1][j + (1 << (i - 1))]);
    }
  }
  auto sparse_table_min = [&](int j, int k) {
    int i = 0;
    while ((1 << (i + 1)) < k - j + 1) {
      i++;
    }
    return sparse_table_combine(sparse_table[i][j],
                                sparse_table[i][k - (1 << i) + 1]);
  };
  auto lowest_common_ancestor = [&](int j, int k) {
    return visit_vertexes[sparse_table_min(
        min(visit_enter[j], visit_enter[k]),
        max(visit_leave[j], visit_leave[k]))];
  };
  auto is_ancestor = [&](int j, int k) {
    return visit_enter[j] >= visit_enter[k] && visit_leave[j] <= visit_leave[k];
  };
  auto distance = [&](int j, int k) {
    if (is_ancestor(j, k)) {
      return abs(visit_depths[visit_enter[j]] - visit_depths[visit_enter[k]]);
    } else {
      int l = lowest_common_ancestor(j, k);
      return abs(visit_depths[visit_enter[j]] - visit_depths[visit_enter[l]]) +
             abs(visit_depths[visit_enter[k]] - visit_depths[visit_enter[l]]);
    }
  };
  auto ask_d = [&](int u) {
    if (u < 0 || u >= n) {
      exit(-2);
    }
    cout << "d"
         << " " << u + 1 << endl;
    int v;
    cin >> v;
    if (v == -1) {
      exit(-3);
    }
    return v;
  };
  auto ask_s = [&](int u) {
    if (u < 0 || u >= n) {
      exit(-4);
    }
    cout << "s"
         << " " << u + 1 << endl;
    int v;
    cin >> v;
    if (v == -1) {
      exit(-5);
    }
    v--;
    return v;
  };
  auto ans = [&](int u) {
    if (u < 0 || u >= n) {
      exit(-6);
    }
    cout << "!"
         << " " << u + 1 << endl;
  };
  int root = 0, last = 0;
  vector<int> candidates((size_t)n);
  iota(candidates.begin(), candidates.end(), 0LL);
  for (int q = 0; (int)candidates.size() > 1; q++) {
    vector<int> new_candidates;
    if (q % 2 == 0) {
      last = root;
      if (q != 0) {
        while (visit_next[last] != -1) {
          last = visit_next[last];
        }
      }
      int d = ask_d(last);
      for (int candidate : candidates) {
        if (distance(candidate, last) == d) {
          new_candidates.push_back(candidate);
        }
      }
    } else {
      root = ask_s(root);
      for (int candidate : candidates) {
        if (is_ancestor(candidate, root)) {
          new_candidates.push_back(candidate);
        }
      }
    }
    candidates = new_candidates;
    if (candidates.empty()) {
      exit(-7);
    }
    root = candidates[0];
    for (int candidate : candidates) {
      root = lowest_common_ancestor(root, candidate);
    }
  }
  ans(root);
  return 0;
}