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 = 2e5 + 5;
vector<int> t[N];
int sz[N], p[N], n;
void init(int u = 1) {
  sz[u] = 1;
  int bestv = -1;
  for (int v : t[u]) {
    if (v == p[u]) continue;
    p[v] = u;
    init(v);
    sz[u] += sz[v];
  }
}
int query(char t, int node) {
  cout << t << ' ' << node << endl;
  int x;
  cin >> x;
  assert(x != -1);
  return x;
}
int find_cent(int u, int tar, int &d) {
  int best = -1, nxt;
  for (int v : t[u]) {
    if (v == p[u]) continue;
    if (sz[v] > best) {
      best = sz[v];
      nxt = v;
    }
  }
  if (best * 2 >= tar) {
    ++d;
    return find_cent(nxt, tar, d);
  } else {
    return u;
  }
}
void rem(int u, int t, int amt) {
  while (1) {
    sz[u] -= amt;
    if (u == t) break;
    u = p[u];
  }
}
int solve(int u, int d1) {
  if (d1 == 0) return u;
  int s1 = query('s', u);
  if (d1 == 1) return s1;
  --d1;
  u = s1;
  while (1) {
    int x = 0;
    int cent = find_cent(u, sz[u], x);
    int d2 = query('d', cent);
    if (d1 - d2 < x) {
      rem(cent, u, sz[cent]);
    } else {
      return solve(cent, d2);
    }
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    t[x].push_back(y);
    t[y].push_back(x);
  }
  init();
  int ans = solve(1, query('d', 1));
  cout << "! " << ans << endl;
}