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>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rnd(chrono::steady_clock().now().time_since_epoch().count());
void answer(int x) {
  cout << "! " << x << "\n";
  fflush(stdout);
  exit(0);
}
map<int, int> qr;
int ask_dist(int x) {
  cout << "d " << x << "\n";
  fflush(stdout);
  int res;
  cin >> res;
  if (res == 0) answer(x);
  return res;
}
int ask_next(int x) {
  cout << "s " << x << "\n";
  fflush(stdout);
  int res;
  cin >> res;
  return res;
}
const int N = (int)2e5 + 9;
vector<int> T[N];
int dis[N];
int pi[N];
void dfs1(int u, int par) {
  pi[u] = par;
  for (auto x : T[u]) {
    if (x != par) {
      dis[x] = dis[u] + 1;
      dfs1(x, u);
    }
  }
}
bool use[N];
int subt[N];
void dfs2(int u, int par) {
  subt[u] = 1;
  for (auto x : T[u]) {
    if (use[x] || x == par) continue;
    dfs2(x, u);
    subt[u] += subt[x];
  }
}
int d0;
void decomp(int node) {
  dfs2(node, -1);
  int sz = subt[node] / 2;
  int pr = -1;
  bool ok = true;
  while (ok) {
    ok = false;
    for (auto x : T[node]) {
      if (use[x] || x == pr) continue;
      if (subt[x] > sz) {
        pr = node;
        node = x;
        ok = true;
        break;
      }
    }
  }
  use[node] = true;
  if (d0 == ask_dist(node) + dis[node]) {
    decomp(ask_next(node));
  } else {
    decomp(pi[node]);
  }
}
int main() {
  int n;
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    T[u].push_back(v);
    T[v].push_back(u);
  }
  dfs1(1, -1);
  d0 = ask_dist(1);
  decomp(1);
  return 0;
}