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;
set<int> node[200000];
int getDepth(int p) {
  cout << "d " << p + 1 << endl;
  int depth;
  cin >> depth;
  return depth;
}
int parent[200000];
int depth[200000];
void setDepth(int maxdepth, int p, int par = -1, int d = 0) {
  depth[p] = d;
  parent[p] = par;
  node[p].erase(par);
  if (d == maxdepth) node[p].clear();
  for (int i : node[p]) setDepth(maxdepth, i, p, d + 1);
}
int sz[200000];
int newroot;
void genSizes(int n, int maxdepth, int p) {
  sz[p] = (depth[p] == maxdepth);
  int good_root = 1;
  for (int i : node[p]) {
    genSizes(n, maxdepth, i);
    sz[p] += sz[i];
    if (sz[i] >= (n + 1) / 2) good_root = 0;
  }
  if (sz[p] < (n + 1) / 2) good_root = 0;
  if (good_root) newroot = p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].insert(b);
    node[b].insert(a);
  }
  int root = 0;
  int depth_x = getDepth(root);
  setDepth(depth_x, root);
  while (1) {
    genSizes(0, depth_x, root);
    newroot = -1;
    genSizes(sz[root], depth_x, root);
    assert(newroot != -1);
    if (sz[root] == 1) {
      cout << "! " << newroot + 1 << endl;
      return 0;
    }
    int d = getDepth(newroot);
    if (d == 0) {
      root = newroot;
      continue;
    }
    if (depth[newroot] + d == depth_x) {
      root = newroot;
      cout << "s " << newroot + 1 << endl;
      int subtree;
      cin >> subtree;
      root = subtree - 1;
    } else {
      node[parent[newroot]].erase(newroot);
    }
  }
}