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 n;
vector<vector<int>> adj;
vector<int> sz, depth;
int centroid(int u, int m) {
  int par = -1;
  bool ch = 1;
  while (ch) {
    ch = 0;
    for (int v : adj[u])
      if (v != par && sz[v] > m / 2) {
        par = u;
        u = v;
        ch = 1;
        break;
      }
  }
  return u;
}
int fillsz(int u, int par = -1) {
  sz[u] = 1;
  for (int v : adj[u])
    if (v != par) {
      sz[u] += fillsz(v, u);
    }
  return sz[u];
}
void filldepth(int u, int par = -1) {
  for (int v : adj[u])
    if (v != par) {
      depth[v] = depth[u] + 1;
      filldepth(v, u);
    }
}
int parent(int u, int x, int par = -1) {
  if (u == x) return par;
  for (int v : adj[u])
    if (v != par) {
      int r = parent(v, x, u);
      if (r != -1) return r;
    }
  return -1;
}
void cut_edge(int u, int v) {
  for (auto it = adj[u].begin(); it != adj[u].end(); ++it)
    if (*it == v) {
      adj[u].erase(it);
      break;
    }
  for (auto it = adj[v].begin(); it != adj[v].end(); ++it)
    if (*it == u) {
      adj[v].erase(it);
      break;
    }
}
int query_dist(int u) {
  cout << "d " << u + 1 << endl;
  int r;
  cin >> r;
  return r;
}
int query_second(int u) {
  cout << "s " << u + 1 << endl;
  int r;
  cin >> r;
  return r - 1;
}
int main() {
  cin >> n;
  adj.resize(n);
  sz.resize(n);
  depth.resize(n);
  for (int i = 0; i < (int)n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a - 1].push_back(b - 1);
    adj[b - 1].push_back(a - 1);
  }
  depth[0] = 0;
  filldepth(0);
  int dx = query_dist(0);
  if (dx == 0) {
    cout << "! 1" << endl;
    return 0;
  }
  int u = 0;
  while (1) {
    if (adj[u].size() == 0) {
      cout << "! " << u + 1 << endl;
      return 0;
    }
    int m = fillsz(u);
    int c = centroid(u, m);
    int dc = query_dist(c);
    if (dc == 0) {
      cout << "! " << c + 1 << endl;
      return 0;
    }
    if (depth[c] + dc == dx) {
      int v = query_second(c);
      cut_edge(c, v);
      u = v;
    } else {
      int p = parent(u, c);
      assert(p != -1);
      cut_edge(c, p);
    }
  }
}