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 MAXN = 2e5 + 7;
const long long MOD = 1e9 + 7;
const long long inf = 1e13;
int n, ans[MAXN], visited[MAXN], sub[MAXN], parent[MAXN], height[MAXN];
vector<int> G[MAXN];
void dfs_pre(int u, int p = -1) {
  for (auto v : G[u])
    if (v != p) height[v] = height[u] + 1, parent[v] = u, dfs_pre(v, u);
}
int dist(int x) {
  cout << "d " << x << endl;
  int ans;
  cin >> ans;
  return ans;
}
int second(int u) {
  cout << "s " << u << endl;
  int ans;
  cin >> ans;
  return ans;
}
void print(int a) {
  cout << "! " << a << endl;
  exit(0);
}
int h1;
void dfs_cent(int u, int p) {
  sub[u] = 1;
  for (auto v : G[u])
    if (!visited[v] && v != p) {
      dfs_cent(v, u);
      sub[u] += sub[v];
    }
}
void CentroidDecompose(int root) {
  dfs_cent(root, -1);
  int c = root, sz = sub[c], par = -1;
  while (1) {
    int p = c;
    for (auto u : G[c])
      if (!visited[u] && sub[u] > sz / 2 && par != u) p = u;
    if (p == c) break;
    par = c;
    c = p;
  }
  int cur = dist(c);
  if (cur == 0) print(c);
  visited[c] = 1;
  if (h1 == height[c] + cur) {
    int x = second(c);
    CentroidDecompose(x);
  } else
    CentroidDecompose(parent[c]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  h1 = dist(1);
  dfs_pre(1);
  CentroidDecompose(1);
}