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 N = (int)2e5 + 7;
int n, root = 1, sub[N], dep[N], dep_me;
bool rem[N];
vector<int> nodes;
vector<int> g[N];
void dfs(int a, int p = 0) {
  sub[a] = 1;
  for (auto &b : g[a]) {
    if (!rem[b] || b == p) continue;
    dep[b] = 1 + dep[a];
    dfs(b, a);
    sub[a] += sub[b];
  }
}
void recalc() {
  nodes.clear();
  dep[root] = 0;
  dfs(root);
}
int find_centroid(int a, int p = 0) {
  int mx = sub[root] - sub[a];
  for (auto &b : g[a]) {
    if (!rem[b] || b == p) continue;
    mx = max(mx, sub[b]);
  }
  if (mx <= sub[root] / 2) {
    return a;
  } else {
    for (auto &b : g[a]) {
      if (sub[b] == mx) {
        return find_centroid(b, a);
      }
    }
  }
  assert(0);
}
map<int, int> memo;
int get_dist(int a) {
  if (memo.count(a)) return memo[a];
  cout << "d " << a << endl;
  cin >> memo[a];
  if (memo[a] == 0) {
    cout << "! " << a << endl;
    exit(0);
  }
  return memo[a];
}
void solve() {
  recalc();
  if (sub[root] == 1) {
    cout << "! " << root << endl;
    exit(0);
  }
  int c = find_centroid(root);
  if (dep[c] + get_dist(c) == dep_me) {
    cout << "s " << c << endl;
    int node;
    cin >> node;
    rem[c] = 0;
    dep_me -= dep[node];
    root = node;
    solve();
    return;
  }
  rem[c] = 0;
  solve();
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) rem[i] = 1, nodes.push_back(i);
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dep_me = get_dist(1);
  solve();
}