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;
const int X = 69;
const int MAXN = 2e5 + 5;
int n;
vector<int> graph[MAXN];
int parent[MAXN];
int depth[MAXN];
int baseDist;
int treeSize[MAXN];
bool wasCentroid[MAXN];
int DFSAnswer1(int x, int last, int dist) {
  if (x == X) return dist;
  for (int y : graph[x]) {
    if (y != last) {
      int val = DFSAnswer1(y, x, dist + 1);
      if (val != -1) return val;
    }
  }
  return -1;
}
int DFSAnswer2(int x, int last, int dist) {
  if (x == X) return x;
  for (int y : graph[x]) {
    if (y != last) {
      int val = DFSAnswer2(y, x, dist + 1);
      if (val != -1) return y;
    }
  }
  return -1;
}
int answerQuery1(int u) { return DFSAnswer1(u, -1, 0); }
int answerQuery2(int u) { return DFSAnswer2(u, parent[u], 0); }
int query1(int u) {
  cout << "d " << u << '\n';
  cout.flush();
  int ans;
  cin >> ans;
  return ans;
}
int query2(int u) {
  cout << "s " << u << '\n';
  cout.flush();
  int ans;
  cin >> ans;
  return ans;
}
void DFSInit1(int x, int last, int level) {
  depth[x] = level;
  parent[x] = last;
  for (int y : graph[x]) {
    if (y != last) {
      DFSInit1(y, x, level + 1);
    }
  }
}
void DFSInit2(int x, int last) {
  treeSize[x] = 1;
  for (int y : graph[x]) {
    if (y == last) continue;
    if (wasCentroid[y] == true) continue;
    DFSInit2(y, x);
    treeSize[x] += treeSize[y];
  }
}
int findCentroid(int x) {
  int root = x;
  bool change = false;
  int last = -1;
  int centroid = x;
  DFSInit2(x, -1);
  do {
    change = false;
    for (int y : graph[centroid]) {
      if (y == last) continue;
      if (wasCentroid[y] == true) continue;
      if (treeSize[y] > treeSize[root] / 2) {
        last = centroid;
        centroid = y;
        change = true;
        break;
      }
    }
  } while (change == true);
  return centroid;
}
bool isAncestor(int x) {
  int dist = query1(x);
  if (dist == 0) {
    cout << "! " << x << '\n';
    cout.flush();
    exit(0);
  }
  if (baseDist == depth[x] + dist) return true;
  return false;
}
int solve(int x) {
  int centroid = findCentroid(x);
  wasCentroid[centroid] = true;
  if (treeSize[x] == 1) return x;
  if (isAncestor(centroid) == true) return solve(query2(centroid));
  return solve(parent[centroid]);
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  DFSInit1(1, -1, 0);
  baseDist = query1(1);
  cout << "! " << solve(1) << '\n';
  cout.flush();
}