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 INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
const int MAX = (int)2e5 + 10;
int queryd(int u) {
  cout << "d " << u + 1 << endl;
  cout.flush();
  int d;
  cin >> d;
  return d;
}
int querys(int u) {
  cout << "s " << u + 1 << endl;
  cout.flush();
  int d;
  cin >> d;
  return d - 1;
}
void resp(int x) {
  cout << "! " << x + 1 << endl;
  cout.flush();
}
int n, root, cent;
vector<int> g[MAX];
bool block[MAX];
int sizet[MAX], dist[MAX];
void get_dist(int u, int p, int d) {
  dist[u] = d;
  for (int e : g[u])
    if (e != p and !block[e]) {
      get_dist(e, u, d + 1);
    }
}
int get_size(int u, int p) {
  sizet[u] = 1;
  for (int e : g[u])
    if (e != p and !block[e]) {
      sizet[u] += get_size(e, u);
    }
  return sizet[u];
}
int centroid(int u, int p, int size) {
  for (int e : g[u])
    if (e != p and !block[e]) {
      if (sizet[e] > size / 2) {
        return centroid(e, u, size);
      }
    }
  return u;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < n; i++) block[i] = false;
  root = 0;
  int dx = queryd(root);
  while (1) {
    if (sizet[root] == 1 or dx == 0) {
      break;
    }
    get_size(root, root);
    cent = centroid(root, root, sizet[root]);
    get_dist(root, root, 0);
    if (cent != root) {
      int dcx = queryd(cent);
      if (dcx == 0) {
        root = cent;
        break;
      }
      if (dx == dcx + dist[cent]) {
        root = querys(cent);
        dx = dcx - 1;
      } else {
        root = querys(root);
        dx--;
      }
    } else {
      root = querys(root);
      dx--;
    }
    block[cent] = true;
  }
  resp(root);
  return 0;
}