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 long long int MOD = 1000000007;
const long long BIG = 1446803456761533460;
const int Big = 336860180;
stringstream sss;
const long long int maxn = 200010;
long long int n;
vector<long long int> g[maxn];
long long int sz[maxn];
long long int ht[maxn];
long long int pars[maxn];
bool vis[maxn];
void dfs_ht(long long int x, long long int par) {
  pars[x] = par;
  for (auto y : g[x]) {
    if (y != par) {
      ht[y] = ht[x] + 1;
      dfs_ht(y, x);
    }
  }
}
void dfs_sz(long long int x, long long int par) {
  sz[x] = 1;
  for (auto y : g[x]) {
    if (y != par && !vis[y]) {
      dfs_sz(y, x);
      sz[x] += sz[y];
    }
  }
}
long long int centroid(long long int x, long long int par, long long int szt) {
  for (auto y : g[x]) {
    if (y != par && !vis[y] && sz[y] * 2 > szt) {
      return centroid(y, x, szt);
    }
  }
  return x;
}
void MAIN() {
  cin >> n;
  for (long long int i = (0); i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    g[a].push_back(b), g[b].push_back(a);
  }
  dfs_ht(1, 0);
  long long int hx;
  cout << "d 1\n";
  cout.flush();
  cin >> hx;
  long long int v = 1;
  while (true) {
    dfs_sz(v, 0);
    if (sz[v] == 1) {
      cout << "! " << v << '\n';
      break;
    }
    long long int c = centroid(v, 0, sz[v]);
    vis[c] = true;
    v = pars[c];
    cout << "d " << c << '\n';
    long long int d;
    cout.flush();
    cin >> d;
    if (d == 0) {
      cout << "! " << c << '\n';
      break;
    }
    if (hx == ht[c] + d) {
      cout << "s " << c << '\n';
      cout.flush();
      cin >> v;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
5
1 2
1 3
3 4
3 5
    )";
  MAIN();
  return 0;
}