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;
using LL = long long;
template <typename T>
void no(const T& s) {
  cout << s;
  exit(0);
}
constexpr int maxn = 240000;
vector<int> G[maxn];
int dep, nxt;
int sz[maxn], f[maxn], d[maxn], ok[maxn];
vector<int> ans;
void DFS1(int u, int fa = 0) {
  f[u] = fa;
  if (d[u] == dep) ok[u] = sz[u] = 1;
  for (int v : G[u])
    if (v != f[u]) {
      d[v] = d[u] + 1;
      DFS1(v, u);
      sz[u] += sz[v];
    }
}
void DFS2(int u) {
  if (sz[u] == 1) {
    nxt = u;
    return;
  }
  int maxv = 0;
  for (int v : G[u])
    if (v != f[u])
      if (sz[v] > sz[maxv]) maxv = v;
  if (maxv) DFS2(maxv);
}
void DFS3(int u, int d, int fa) {
  if (d) ok[u] = 0;
  for (int v : G[u])
    if (v != fa) DFS3(v, d - 1, u);
}
void DFS4(int u) {
  sz[u] = ok[u];
  for (int v : G[u])
    if (f[u] != v) {
      DFS4(v);
      sz[u] += sz[v];
    }
}
void DFS5(int u, int x) {
  if (u == x) return;
  ok[u] = 0;
  for (int v : G[u])
    if (f[u] != v) {
      DFS5(v, x);
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i < n; i += 1) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  cout << "d 1" << endl;
  cin >> dep;
  DFS1(1);
  for (int i = 1; i <= n; i += 1)
    if (ok[i]) ans.push_back(i);
  int u = 1;
  while (true) {
    if (ans.size() == 1) {
      cout << "! " << ans[0];
      return 0;
    }
    if (d[u] == dep) {
      cout << "! " << u;
      return 0;
    }
    int maxv = 0, cnt = 0;
    for (int v : G[u])
      if (v != f[u]) {
        if (sz[v] > sz[maxv]) maxv = v;
        if (sz[v]) cnt += 1;
      }
    if (cnt == 1) {
      u = maxv;
      continue;
    }
    if (sz[maxv] <= sz[u] / 2) {
      cout << "s " << u << endl;
      cin >> u;
      DFS5(1, u);
      DFS4(1);
      continue;
    }
    DFS2(maxv);
    cout << "d " << nxt << endl;
    int dd;
    cin >> dd;
    DFS3(nxt, dd, 0);
    DFS4(1);
    ans.clear();
    for (int i = 1; i <= n; i += 1)
      if (ok[i]) ans.push_back(i);
  }
  return 0;
}