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;
int n, dp[200005], par[200005], at, si, bi, be, cd, ld = -1, dis[200005];
bool g[200005];
vector<int> v[200005];
int ta, tb;
void dfs(int a, int b) {
  si++;
  for (int i : v[a])
    if (i != b && !g[i]) dfs(i, a);
}
void ddfs(int a, int b) {
  dp[a] = 1;
  dis[a] = dis[b] + 1;
  int ma = 0;
  for (int i : v[a])
    if (i != b && !g[i]) {
      ddfs(i, a);
      dp[a] += dp[i];
      ma = max(ma, dp[i]);
    }
  ma = max(ma, si - dp[a]);
  if (ma < be) {
    be = ma;
    bi = a;
  }
}
void dff(int a, int b) {
  par[a] = b;
  for (int i : v[a])
    if (i != b) dff(i, a);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &ta, &tb);
    v[ta].push_back(tb);
    v[tb].push_back(ta);
  }
  dff(1, 0);
  at = 1;
  si = n;
  while (1) {
    if (si == 1 || ld == 1) {
      printf("! %d\n", at);
      fflush(stdout);
      return 0;
    }
    be = INT_MAX;
    si = 0;
    dfs(at, 0);
    ddfs(at, 0);
    printf("d %d\n", bi);
    fflush(stdout);
    scanf("%d", &cd);
    if (cd == 0) {
      printf("! %d\n", bi);
      fflush(stdout);
      return 0;
    }
    if (ld == -1)
      if (bi == 1) {
        puts("s 1");
        fflush(stdout);
        g[1] = 1;
        ld = cd;
        scanf("%d", &at);
      } else {
        printf("d %d\n", par[bi]);
        fflush(stdout);
        int tmp;
        scanf("%d", &tmp);
        if (tmp == 0) {
          printf("! %d\n", par[bi]);
          fflush(stdout);
          return 0;
        }
        if (tmp < cd) {
          g[bi] = 1;
          at = par[bi];
        } else {
          printf("s %d\n", bi);
          fflush(stdout);
          g[bi] = 1;
          ld = cd;
          scanf("%d", &at);
        }
      }
    else if (cd + dis[bi] == ld) {
      printf("s %d\n", bi);
      fflush(stdout);
      g[bi] = 1;
      ld = cd;
      scanf("%d", &at);
    } else
      g[bi] = 1;
  }
}