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 int N = 2e5 + 5;
vector<int> adj[N];
int sum1[N], now1, wow, cac1, level[N];
bool used[N];
void dfs(int x, int p) {
  if (level[x] == now1) {
    sum1[x]++;
  }
  for (int i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p) {
      level[adj[x][i]] = level[x] + 1;
      dfs(adj[x][i], x);
      sum1[x] += sum1[adj[x][i]];
    }
  }
}
void dfs1(int x, int p) {
  sum1[x] = 0;
  if (level[x] == now1) {
    sum1[x]++;
  }
  for (int i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p && !used[adj[x][i]]) {
      dfs1(adj[x][i], x);
      sum1[x] += sum1[adj[x][i]];
    }
  }
}
int findcen(int x, int p, int sz) {
  for (int i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p && sum1[adj[x][i]] > sz / 2 && !used[adj[x][i]]) {
      return findcen(adj[x][i], x, sz);
    }
  }
  return x;
}
void findans(int x, int p, int sz) {
  if (sz == 1) {
    cac1 = x;
    return;
  }
  dfs1(x, x);
  int mid = findcen(x, x, sz);
  used[mid] = true;
  cout << "d " << mid << endl;
  fflush(stdout);
  cin >> wow;
  if (wow + level[mid] == now1) {
    if (wow == 0) {
      cout << "! " << mid << endl;
      fflush(stdout);
      exit(0);
    }
    cout << "s " << mid << endl;
    fflush(stdout);
    cin >> wow;
    findans(wow, wow, sum1[wow]);
  } else {
    findans(x, x, sz - sum1[mid]);
  }
}
int dfs2(int x, int p) {
  if (level[x] == now1) {
    return x;
  }
  int i, j;
  for (i = 0; i < adj[x].size(); i++) {
    if (adj[x][i] != p && !used[adj[x][i]]) {
      j = dfs2(adj[x][i], x);
      if (j != -1) {
        return j;
      }
    }
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, i, j, k;
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> j >> k;
    adj[j].push_back(k);
    adj[k].push_back(j);
  }
  cout << "d 1" << endl;
  fflush(stdout);
  cin >> now1;
  dfs(1, 1);
  findans(1, 1, sum1[1]);
  i = dfs2(cac1, cac1);
  cout << "! " << i << endl;
  fflush(stdout);
}