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;
vector<vector<int> > G;
int fa[200010], sze[200010], deep[200010], son[200010], bigson[200010], depx;
void pre(int u) {
  sze[u] = 1;
  son[u] = 0;
  for (auto v : G[u])
    if (v != fa[u]) {
      fa[v] = u;
      deep[v] = deep[u] + 1;
      pre(v);
      sze[u] += sze[v];
      if (!son[u] || sze[v] >= sze[son[u]]) {
        son[u] = v;
      }
    }
}
void f5() { fflush(stdout); }
void solve(int u, int dxu) {
  int v, dxv, y;
  if (dxu == -1) {
    printf("d %d\n", u);
    f5();
    scanf("%d", &dxu);
  }
  if (dxu == 0) {
    printf("! %d\n", u);
    f5();
    return;
  }
  if (u == 1) {
    depx = dxu;
  }
  if (!bigson[u]) {
    printf("s %d\n", u);
    f5();
    scanf("%d", &v);
    solve(v, dxu - 1);
  } else {
    v = u;
    for (int i = 1; i <= dxu && son[v]; ++i) {
      v = son[v];
    }
    printf("d %d\n", v);
    f5();
    scanf("%d", &dxv);
    if (dxv == 0) {
      printf("! %d\n", v);
      f5();
      return;
    } else {
      y = u;
      for (int i = 0; i < dxu; ++i) {
        if (deep[v] + depx - 2 * deep[y] == dxv) {
          break;
        }
        y = son[y];
      }
      printf("s %d\n", y);
      f5();
      scanf("%d", &y);
      solve(y, depx - deep[y]);
    }
  }
}
int main() {
  while (scanf("%d", &n) != EOF) {
    G.clear();
    G.resize(n + 1);
    for (int i = 2, u, v; i <= n; ++i) {
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    son[0] = 0;
    sze[0] = 0;
    deep[1] = 0;
    pre(1);
    memset(bigson, 0, sizeof bigson);
    for (int i = 1; i <= n; ++i)
      if (son[i]) {
        bigson[son[i]] = 1;
      }
    solve(1, -1);
  }
  return 0;
}