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 double eps = 1e-8;
const long long inf = 1e9;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 10;
struct node {
  int d;
  node *to;
} * e[maxn];
char str[20];
int siz[maxn], fa[maxn], dep[maxn], minsiz, root, pre_root, nn;
bool hav[maxn];
void dfs(int d) {
  int maxs = 0;
  node *p = e[d];
  siz[d] = 1;
  while (p) {
    if (fa[d] != p->d && !hav[p->d]) {
      fa[p->d] = d;
      dep[p->d] = dep[d] + 1;
      dfs(p->d);
      siz[d] += siz[p->d];
      maxs = max(maxs, siz[p->d]);
    }
    p = p->to;
  }
  maxs = max(maxs, nn - siz[d]);
  if (maxs < minsiz) minsiz = maxs, root = d;
}
int main() {
  node *p;
  int n, x, y, i, r, dist;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    p = new node();
    p->d = y;
    p->to = e[x];
    e[x] = p;
    p = new node();
    p->d = x;
    p->to = e[y];
    e[y] = p;
  }
  gets(str);
  printf("d %d\n", 1);
  fflush(stdout);
  scanf("%d", &dist);
  if (dist == 0) {
    printf("! %d\n", 1);
    fflush(stdout);
    return 0;
  }
  nn = n;
  root = 1;
  while (1) {
    pre_root = root;
    dep[pre_root] = 0;
    minsiz = inf;
    dfs(root);
    if (root == pre_root) {
      printf("s %d\n", root);
      fflush(stdout);
      hav[root] = 1;
      scanf("%d", &root);
      dist--;
      if (dist == 0) {
        printf("! %d\n", root);
        fflush(stdout);
        return 0;
      }
      nn = siz[root];
    } else {
      printf("d %d\n", root);
      fflush(stdout);
      scanf("%d", &r);
      if (r == 0) {
        printf("! %d\n", root);
        fflush(stdout);
        return 0;
      }
      hav[root] = 1;
      if (r + dep[root] == dist) {
        printf("s %d\n", root);
        fflush(stdout);
        dist -= dep[root] + 1;
        scanf("%d", &root);
        if (dist == 0) {
          printf("! %d\n", root);
          fflush(stdout);
          return 0;
        }
        nn = siz[root];
      } else {
        nn = siz[pre_root] - siz[root];
        root = pre_root;
      }
    }
  }
  return 0;
}