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;
vector<int> V[200005];
int deep[200005];
int fa[200005];
int sz[200005];
bool g[200005];
int depth = 0;
int now, now_root;
void dfs1(int x, int f) {
  sz[x] = 0;
  if (deep[x] == depth) sz[x] = 1;
  for (auto v : V[x]) {
    if (v != f) {
      deep[v] = deep[x] + 1;
      dfs1(v, x);
      fa[v] = x;
      sz[x] = sz[x] + sz[v];
    }
  }
}
int dfs2(int x) {
  for (auto v : V[x]) {
    if (v == fa[x]) continue;
    if (sz[v] > now / 2) return dfs2(v);
  }
  return x;
}
void dfs3(int x) {
  sz[x] = 0;
  if (deep[x] == depth && g[x]) sz[x] = 1;
  for (auto v : V[x]) {
    if (v == fa[x]) continue;
    dfs3(v);
    sz[x] += sz[v];
  }
}
void dfs4(int x) {
  if (deep[x] == depth) g[x] = false;
  for (auto v : V[x]) {
    if (v == fa[x]) continue;
    dfs4(v);
  }
}
int main() {
  int n;
  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);
  }
  for (int i = 1; i <= n; i++) g[i] = true;
  deep[1] = 0;
  printf("d 1\n");
  fflush(stdout);
  scanf("%d", &depth);
  dfs1(1, -1);
  now = sz[1];
  now_root = 1;
  while (now >= 1) {
    if (deep[now_root] == depth) {
      printf("! %d\n", now_root);
      fflush(stdout);
      return 0;
    }
    int t = dfs2(now_root);
    if (t == now_root) {
      printf("s %d\n", t);
      fflush(stdout);
      int tt;
      scanf("%d", &tt);
      dfs3(tt);
      now = sz[tt];
      now_root = tt;
    } else {
      printf("d %d\n", t);
      fflush(stdout);
      int tt;
      scanf("%d", &tt);
      if (tt == depth - deep[t]) {
        if (tt == 0) {
          printf("! %d\n", t);
          fflush(stdout);
          return 0;
        }
        printf("s %d\n", t);
        fflush(stdout);
        int ttt;
        scanf("%d", &ttt);
        dfs3(ttt);
        now = sz[ttt];
        now_root = ttt;
      } else {
        dfs4(t);
        dfs3(now_root);
        now = sz[now_root];
      }
    }
  }
  return 0;
}