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 MOD = 1000000007;
const double pi = atan(1.0) * 4.0;
const double eps = 1e-8;
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
int bc(int n) { return n ? bc((n - 1) & n) + 1 : 0; }
const int N = 1000005;
int i, j, k, m, n;
int ans;
vector<int> adj[N];
int dead[N], sz[N], h[N], pr[N];
int dist(int u) {
  cout << "d " << u << endl;
  int ret;
  cin >> ret;
  return ret;
}
int sub(int u) {
  cout << "s " << u << endl;
  int ret;
  cin >> ret;
  return ret;
}
void over(int u) {
  cout << "! " << u << endl;
  exit(0);
}
void dfs(int i, int par) {
  sz[i] = 1;
  for (int j : adj[i])
    if (j != par && !dead[j]) {
      pr[j] = i;
      h[j] = h[i] + 1;
      dfs(j, i);
      sz[i] += sz[j];
    }
}
int main() {
  int tn = 1;
  for (int tt = 1; tt <= tn; tt++) {
    cin >> n;
    for (int k = 0; k < n - 1; k++) {
      cin >> i >> j;
      adj[i].push_back(j);
      adj[j].push_back(i);
    }
    int d = dist(1);
    int root = 1;
    while (1) {
      if (d == 0) over(root);
      h[root] = 0;
      dfs(root, -1);
      int oc = -1;
      for (int j : adj[root])
        if (!dead[j]) {
          if (oc == -1)
            oc = j;
          else
            oc = -2;
        }
      if (oc > 0) {
        root = oc;
        d--;
        continue;
      }
      int y = root;
      int sg = 1;
      while (sg) {
        sg = 0;
        for (int j : adj[y])
          if (!dead[j] && h[j] == h[y] + 1 && 2 * sz[j] + 1 >= sz[root]) {
            y = j;
            sg = 1;
            break;
          }
      }
      if (y == root) {
        dead[root] = 1;
        root = sub(y);
        d--;
      } else {
        int d2 = dist(y);
        if (d2 == 0) over(y);
        if (h[y] + d2 == d) {
          root = y;
          dead[pr[root]] = 1;
          d = d2;
        } else {
          dead[y] = 1;
        }
      }
    }
  }
  return 0;
}