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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
const int N = 3e5 + 10;
const int M = 22;
int n;
int P[N][M];
int big[N][M], low[N][M], level[N];
vector<int> graph[N];
int qd(int x) {
  cout << "d " << x << endl;
  cin >> x;
  return x;
}
int qs(int x) {
  cout << "s " << x << endl;
  cin >> x;
  return x;
}
void ANS(int x) {
  cout << "! " << x << endl;
  exit(0);
}
void dfs(int u, int last, int l) {
  level[u] = l;
  P[u][0] = last;
  for (int v : graph[u])
    if (v != last) dfs(v, u, l + 1);
}
void process() {
  dfs(1, 1, 0);
  for (int j = 1; j < M; j++)
    for (int i = 1; i <= n; i++) {
      P[i][j] = P[P[i][j - 1]][j - 1];
    }
}
int lca(int u, int v) {
  if (u == v) return u;
  if (level[u] < level[v]) swap(u, v);
  for (int i = M - 1; i >= 0; i--)
    if (level[u] - (1 << i) >= level[v]) u = P[u][i];
  if (u == v) return u;
  for (int i = M - 1; i >= 0; i--) {
    if (P[u][i] != P[v][i]) u = P[u][i], v = P[v][i];
  }
  return P[u][0];
}
int poder[N];
int prep(int u, int pai) {
  for (int v : graph[u])
    if (v != pai) poder[u] += prep(v, u);
  return poder[u];
}
void prepre(vector<int> &caras) {
  memset(poder, 0, sizeof(poder));
  for (int x : caras) poder[x] = 1;
  prep(1, 1);
}
void itera(vector<int> &caras) {
  prepre(caras);
  int root = caras[0];
  for (int x : caras) root = lca(root, x);
  int foi = 1;
  while (foi) {
    foi = 0;
    for (int v : graph[root])
      if (v != P[root][0]) {
        if (poder[v] > caras.size() / 2) {
          foi = 1;
          root = v;
          break;
        }
      }
  }
  int d = qd(root);
  int era = level[caras[0]] - level[root];
  if (d == era) {
    root = qs(root);
    vector<int> cara2;
    for (int x : caras)
      if (lca(x, root) == root) cara2.push_back(x);
    caras = cara2;
  } else {
    assert(d > era);
    vector<int> cara2;
    for (int x : caras)
      if (lca(x, root) != root) cara2.push_back(x);
    caras = cara2;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  process();
  int nivel = qd(1);
  if (nivel == 0) ANS(1);
  vector<int> caras;
  for (int i = 1; i <= n; i++)
    if (level[i] == nivel) caras.push_back(i);
  while (caras.size() != 1) {
    itera(caras);
  }
  ANS(caras[0]);
}