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;
long long n, col, cur;
vector<vector<long long>> g;
vector<bool> dead;
vector<long long> s, pred, pred2, first_d;
void dfs0(long long v, long long p) {
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to != p) {
      first_d[to] = first_d[v] + 1;
      dfs0(to, v);
    }
  }
}
void dfs(long long v, long long p) {
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to != p) {
      pred[to] = v;
      dfs(to, v);
    }
  }
}
void dfs2(long long v, long long p) {
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (!dead[to] && to != p) {
      pred2[to] = v;
      dfs2(to, v);
    }
  }
}
long long fs(long long v, long long p) {
  s[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to != p && !dead[to]) {
      s[v] += fs(to, v);
    }
  }
  return s[v];
}
long long fc(long long v, long long p) {
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (!dead[to] && to != p) {
      if (s[to] > col / 2) {
        return fc(to, v);
      }
    }
  }
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  cur = 0;
  col = n;
  g.resize(n);
  dead.resize(n);
  s.resize(n);
  pred.resize(n);
  pred2.resize(n);
  first_d.resize(n);
  for (int i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0, -1);
  first_d[0] = 0;
  dfs0(0, -1);
  long long kek;
  cout << "d " << 1 << endl;
  cin >> kek;
  for (int j = 0; j < 36; j++) {
    if (col == 1) {
      cout << "! " << cur + 1 << endl;
      return 0;
    }
    pred2[cur] = -1;
    dfs2(cur, -1);
    fs(cur, -1);
    long long c = fc(cur, -1);
    long long mem;
    cout << "d " << c + 1 << endl;
    cin >> mem;
    if (mem == 0) {
      cout << "! " << c + 1 << endl;
      return 0;
    }
    if (mem + first_d[c] == kek) {
      cout << "s " << c + 1 << endl;
      long long lol;
      cin >> lol;
      lol--;
      dead[c] = true;
      cur = lol;
      if (lol != pred2[c]) {
        col = s[lol];
      } else {
        col -= s[c];
      }
    } else {
      dead[c] = true;
      cur = pred[c];
      if (pred[c] == pred2[c]) {
        col -= s[c];
      } else {
        col = s[pred[c]];
      }
    }
  }
  return 0;
}