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 n, d;
vector<int> adj[200010];
int par[200010];
int dd[200010];
int sz[200010];
int vis[200010];
void dfs(int v, int p) {
  for (int e : adj[v]) {
    if (e != p) {
      par[e] = v;
      dd[e] = dd[v] + 1;
      dfs(e, v);
    }
  }
}
int dfs_sz(int v, int p) {
  sz[v] = 1;
  for (int e : adj[v]) {
    if (e != p && !vis[e]) {
      sz[v] += dfs_sz(e, v);
    }
  }
  return sz[v];
}
int dfs_root(int v, int p, int n) {
  for (int e : adj[v]) {
    if (e != p && !vis[e] && 2 * sz[e] > n) {
      return dfs_root(e, v, n);
    }
  }
  return v;
}
bool check(int v, int p, int val) {
  if (v == val) return true;
  for (int e : adj[v]) {
    if (e != p && !vis[e] && check(e, v, val)) return true;
  }
  return false;
}
void centroid(int v) {
  dfs_sz(v, v);
  int c = dfs_root(v, v, sz[v]);
  vis[c] = true;
  cout << "d " << c << endl;
  int dist;
  cin >> dist;
  if (dist == 0) {
    cout << "! " << c << endl;
  } else if (dist + dd[c] == d) {
    cout << "s " << c << endl;
    int nxt;
    cin >> nxt;
    for (int e : adj[c]) {
      if (!vis[e] && e != par[c]) {
        if (check(e, c, nxt)) {
          centroid(e);
        }
      }
    }
  } else {
    centroid(par[c]);
  }
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  cout << "d 1" << endl;
  cin >> d;
  dfs(1, 0);
  centroid(1);
  return 0;
}