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;
int n, dep_x;
int ch[N], d[N], par[N];
bool h[N];
vector<int> v[N];
int ask_d(int x) {
  printf("d %d\n", x);
  fflush(stdout);
  int v;
  scanf("%d", &v);
  if (!v) {
    printf("! %d\n", x);
    exit(0);
  }
  return v;
}
int ask_s(int x) {
  printf("s %d\n", x);
  fflush(stdout);
  int v;
  scanf("%d", &v);
  return v;
}
void dfs2(int p, int x) {
  ch[x] = 1;
  for (auto u : v[x]) {
    if (u != p and !h[u]) {
      dfs2(x, u);
      ch[x] += ch[u];
    }
  }
}
void dfs3(int p, int x) {
  d[x] = d[p] + 1;
  par[x] = p;
  for (auto u : v[x]) {
    if (u != p and !h[u]) {
      dfs3(x, u);
    }
  }
}
int find(int p, int x, int t) {
  for (auto u : v[x]) {
    if (u != p and !h[u] and ch[u] * 2 >= t) {
      return find(x, u, t);
    }
  }
  return x;
}
void dfs(int x) {
  dfs2(0, x);
  x = find(0, x, ch[x]);
  h[x] = 1;
  int dist = ask_d(x);
  if (d[x] + dist != dep_x) {
    dfs(par[x]);
  } else {
    dfs(ask_s(x));
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dep_x = ask_d(1);
  d[0] = -1;
  dfs3(0, 1);
  dfs(1);
  return 0;
}