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;
int n, m, k, x, y, z;
int b[300500], c[300500];
int a[300500];
int Abs(int x) { return (x > 0 ? x : -x); }
vector<int> g[200500], gb[200500];
int centroid, tree_size;
int ans(int x);
int askd(int x) {
  cout << "d " << x << endl;
  fflush(stdout);
  int y;
  cin >> y;
  if (y == 0) {
    ans(x);
  }
  return y;
}
int asks(int x) {
  cout << "s " << x << endl;
  fflush(stdout);
  int y;
  cin >> y;
  return y;
}
int ans(int x) {
  cout << "! " << x << endl;
  exit(0);
}
void dfs1(int v, int p = -1) {
  for (auto to : gb[v]) {
    if (to == p) {
      continue;
    }
    dfs1(to, v);
    g[v].push_back(to);
  }
}
void dfs(int v, int lv = 0) {
  b[v] = 0;
  a[v] = 1;
  c[v] = lv;
  for (auto to : g[v]) {
    dfs(to, lv + 1);
    b[v] = max(b[v], a[to]);
    a[v] += a[to];
  }
  if (a[v] > tree_size / 2 && b[v] <= tree_size / 2) {
    centroid = v;
  }
}
void cut_tree(int v, int d) {
  if (g[v].empty()) {
    ans(v);
  }
  dfs(v);
  int dc = askd(centroid);
  if (dc + c[centroid] == d) {
    int vc = asks(centroid);
    tree_size = a[vc];
    if (dc == 1) {
      ans(vc);
    }
    cut_tree(vc, dc - 1);
  } else {
    g[centroid].clear();
    tree_size -= a[centroid] - 1;
    cut_tree(v, d);
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &x, &y);
    gb[x].push_back(y);
    gb[y].push_back(x);
  }
  dfs1(1);
  int d = askd(1);
  tree_size = n;
  cut_tree(1, d);
  return 0;
}