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;
vector<vector<int> > g;
vector<int> sz;
vector<int> dep, p;
int d;
void dfs1(int cur, int par, int depth, int tar) {
  sz[cur] = 0;
  dep[cur] = depth;
  p[cur] = par;
  if (depth > tar) return;
  if (depth == tar) {
    sz[cur] = 1;
    return;
  }
  for (int u : g[cur]) {
    if (u == par) continue;
    dfs1(u, cur, depth + 1, tar);
    sz[cur] += sz[u];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int qc = 1;
  g.resize(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);
  }
  sz.resize(n);
  dep.resize(n);
  p.resize(n);
  cout << "d " << 1 << endl;
  cin >> d;
  if (d == 0) {
    cout << "! 1" << endl;
    return 0;
  }
  dfs1(0, -1, 0, d);
  int cur = 0;
  while (dep[cur] < d) {
    int v = cur;
    while (dep[v] < d) {
      int best = -1;
      assert(sz[v] > 0);
      for (int u : g[v]) {
        if (u == p[v]) continue;
        if (best == -1 || sz[u] > sz[best]) best = u;
      }
      v = best;
    }
    qc++;
    assert(qc <= 36);
    int dis;
    cout << "d " << v + 1 << endl;
    cin >> dis;
    if (dis == 0) {
      cur = v;
      break;
    }
    assert(dis != -1);
    int lca = v;
    for (int i = 0; i < dis / 2; i++) {
      if (lca == -1) return 4;
      lca = p[lca];
    }
    qc++;
    assert(qc <= 36);
    cout << "s " << lca + 1 << endl;
    cin >> cur;
    assert(cur != -1);
    cur--;
  }
  cout << "! " << cur + 1 << endl;
}