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 = 1e9 + 7;
const int MAXN = 2e5 + 10;
int n, dx;
int chain_id[MAXN], ssize[MAXN], parent[MAXN], depth[MAXN];
vector<int> v[MAXN], chain[MAXN];
void dfs(int node, int dad) {
  ssize[node] = 1;
  parent[node] = dad;
  for (int nxt : v[node]) {
    if (nxt == dad) continue;
    depth[nxt] = depth[node] + 1;
    dfs(nxt, node);
    ssize[node] += ssize[nxt];
  }
}
void hld(int node, int dad, int cid) {
  chain_id[node] = cid;
  chain[cid].push_back(node);
  int maks = -1;
  for (int nxt : v[node]) {
    if (nxt == dad) continue;
    if (maks == -1 || ssize[nxt] > ssize[maks]) maks = nxt;
  }
  if (maks == -1) return;
  hld(maks, node, cid);
  for (int nxt : v[node]) {
    if (nxt == dad || nxt == maks) continue;
    hld(nxt, node, nxt);
  }
}
int get_distance(int x) {
  int ret;
  printf("d %d\n", x + 1);
  fflush(stdout);
  scanf("%d", &ret);
  return ret;
}
int get_next(int x) {
  int ret;
  printf("s %d\n", x + 1);
  fflush(stdout);
  scanf("%d", &ret);
  return ret - 1;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(0, -1);
  hld(0, -1, 0);
  dx = get_distance(0);
  int root = 0, dist = dx;
  while (dist > 0) {
    int nxt = get_next(root);
    if (chain_id[nxt] != chain_id[root]) {
      root = nxt;
      --dist;
      continue;
    }
    int B = chain[chain_id[root]].back();
    int dist_b = get_distance(B);
    int dy = (depth[B] + dx - dist_b) / 2;
    dist -= dy - depth[root];
    root = chain[chain_id[root]][dy - depth[root]];
  }
  printf("! %d\n", root + 1);
  fflush(stdout);
  return 0;
}