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 maxn = 2e5 + 5;
int n, len;
vector<int> way[maxn];
int h[maxn], leaf[maxn];
int have, cen;
int cnt[maxn], big[maxn];
int ban[maxn];
void dfs(int u, int last) {
  h[u] = h[last] + 1;
  if (h[u] == len + 1) leaf[u] = 1, have++;
  for (auto v : way[u])
    if (v != last) dfs(v, u);
}
void build(int u, int last) {
  cnt[u] = leaf[u];
  big[u] = 0;
  for (auto v : way[u]) {
    if (v == last || h[u] >= h[v] || ban[v]) continue;
    build(v, u);
    big[u] = max(big[u], cnt[v]);
    cnt[u] += cnt[v];
  }
  big[u] = max(big[u], have - cnt[u]);
  if (!leaf[u] && big[u] <= have / 2 + 1) cen = u;
}
int get(int u, int last) {
  if (leaf[u]) return u;
  for (auto v : way[u]) {
    if (v == last || h[u] >= h[v] || ban[v]) continue;
    int tmp = get(v, u);
    if (tmp) return tmp;
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    way[u].push_back(v);
    way[v].push_back(u);
  }
  int CNT = 0;
  printf("d 1\n");
  fflush(stdout);
  CNT++;
  scanf("%d", &len);
  dfs(1, 0);
  int root = 1, last = 0;
  build(root, last);
  while (have > 1) {
    if (have == 2) {
      int tmp = get(root, last);
      printf("d %d\n", tmp);
      fflush(stdout);
      CNT++;
      int res;
      scanf("%d", &res);
      if (res != 0) ban[tmp] = 1;
      break;
    }
    int tmp = get(cen, 0);
    printf("d %d\n", tmp);
    fflush(stdout);
    CNT++;
    int res;
    scanf("%d", &res);
    if (res <= (h[tmp] - h[cen]) * 2) {
      printf("s %d\n", cen);
      fflush(stdout);
      CNT++;
      scanf("%d", &root);
      last = cen;
    } else
      ban[cen] = 1;
    build(root, last);
    have = cnt[root];
  }
  printf("! %d\n", get(root, last));
  fflush(stdout);
}