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 queryD(int node) {
  cout << "d " << node << "\n";
  fflush(stdin);
  int ans;
  cin >> ans;
  return ans;
}
int second[200002];
int queryS(int node) {
  if (second[node] != 0) return second[node];
  cout << "s " << node << "\n";
  fflush(stdin);
  int ans;
  cin >> ans;
  return second[node] = ans;
}
void answer(int node) {
  cout << "! " << node << "\n";
  fflush(stdout);
}
int n;
int droot;
vector<int> edges[200002];
bool visited[200002];
vector<int> sons[200002];
int level[200002];
int cnt[200002];
int total;
void dfs(int node) {
  visited[node] = true;
  for (int u : edges[node])
    if (visited[u] == false) {
      level[u] = level[node] + 1;
      dfs(u);
      sons[node].push_back(u);
    }
}
bool out[200002];
void compute(int node) {
  cnt[node] = 0;
  if (level[node] == droot) {
    cnt[node] = 1;
    total++;
  }
  for (int u : sons[node])
    if (out[u] == false) {
      compute(u);
      cnt[node] += cnt[u];
    }
}
int root = 1;
bool solve(int node) {
  if (level[node] == droot && total == 1) {
    answer(node);
    return true;
  }
  if (cnt[node] * 2 <= total) {
    int d = queryD(node);
    if (d == 0) {
      answer(node);
      return true;
    }
    if (d + level[node] > droot) {
      out[node] = true;
      total = 0;
      compute(root);
    } else {
      root = node;
      total -= cnt[node];
    }
    return false;
  }
  if (sons[node].size() == 1 && node == root) root = sons[node][0];
  if (node == root && sons[node].size() > 30) {
    int u = queryS(node);
    return solve(u);
  }
  int umx = 0;
  for (int u : sons[node])
    if (out[u] == false) {
      if (cnt[u] > cnt[umx]) umx = u;
    }
  if (umx == 0) return true;
  return solve(umx);
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  dfs(1);
  droot = queryD(1);
  if (droot == 0) {
    answer(1);
    return 0;
  }
  if (sons[1].size() == n - 1) {
    answer(queryS(1));
    return 0;
  }
  compute(1);
  while (solve(root) == false)
    ;
  return 0;
}