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 L = 2e5;
vector<vector<int>> g(L);
vector<int> dp(L), d(L, 0), par(L, -1);
vector<bool> used(L);
int dist;
int ask1(int u) {
  cout << "d " << u + 1 << endl;
  int t;
  cin >> t;
  if (t == -1) {
    while (true) {
    }
  }
  return t;
}
int ask2(int u) {
  cout << "s " << u + 1 << endl;
  int t;
  cin >> t;
  --t;
  if (t == -1) {
    while (true) {
    }
  }
  return t;
}
void dfs(int u) {
  used[u] = true;
  for (auto v : g[u]) {
    if (!used[v]) {
      d[v] = d[u] + 1;
      par[v] = u;
      dfs(v);
      dp[u] += dp[v];
    }
  }
  if (d[u] == dist) dp[u]++;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dist = ask1(0);
  dfs(0);
  int q = 0;
  if (dist == 0) {
    cout << "! 1" << endl;
    return 0;
  } else {
    while (d[q] != dist) {
      int r = q;
      while (d[r] != dist) {
        int t = -1;
        int ma = 0;
        for (auto v : g[r]) {
          if (d[v] > d[r]) {
            if (dp[v] > ma) {
              ma = dp[v];
              t = v;
            }
          }
        }
        r = t;
      }
      int newdist = ask1(r);
      if (newdist == 0) {
        cout << "! " << r + 1 << endl;
        return 0;
      } else {
        while (d[r] * 2 != 2 * dist - newdist) {
          r = par[r];
        }
        q = ask2(r);
      }
    }
    cout << "! " << q + 1 << endl;
    return 0;
  }
  cin >> n;
}