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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int n, lvl[200009], par[200009];
vector<int> adj[200009];
int LCA(int x, int y) {
  while (x != y) {
    if (lvl[x] < lvl[y]) swap(x, y);
    x = par[x];
  }
  return x;
}
int dis(int x, int y) { return lvl[x] + lvl[y] - 2 * lvl[LCA(x, y)]; }
int ask, X = 6;
int d(int x) {
  ask++;
  assert(1 <= x and x <= n);
  cout << "d " << x << endl;
  int res;
  cin >> res;
  return res;
}
bool ata(int x) {
  if (lvl[X] <= lvl[x]) return 0;
  int tmp = X;
  while (lvl[tmp] > lvl[x]) tmp = par[tmp];
  return (tmp == x);
}
int tap(int x) {
  int mn = 1000000007, who = -1;
  for (__typeof((adj[x]).begin()) it = (adj[x]).begin(); it != (adj[x]).end();
       it++)
    if (umin(mn, dis(*it, X))) who = *it;
  return who;
}
int s(int x) {
  ask++;
  assert(1 <= x and x <= n);
  cout << "s " << x << endl;
  int res;
  cin >> res;
  return res;
}
void dfs(int nd, int pr) {
  for (__typeof((adj[nd]).begin()) it = (adj[nd]).begin();
       it != (adj[nd]).end(); it++)
    if (*it != pr) {
      lvl[*it] = lvl[nd] + 1;
      par[*it] = nd;
      dfs(*it, nd);
    }
}
int sub[200009], vis[200009];
void f1(int nd, int pr) {
  sub[nd] = 1;
  for (int i = 0; i < int(adj[nd].size()); i++) {
    int to = adj[nd][i];
    if (to == pr or vis[to]) continue;
    f1(to, nd);
    sub[nd] += sub[to];
  }
}
int f2(int nd, int pr, int sz) {
  for (int i = 0; i < int(adj[nd].size()); i++) {
    int to = adj[nd][i];
    if (to == pr or vis[to]) continue;
    if (sub[to] > sz) return f2(to, nd, sz);
  }
  return nd;
}
int get(int nd) {
  f1(nd, -1);
  return f2(nd, -1, sub[nd] >> 1);
}
int ana(int x, int y) {
  while (y--) x = par[x];
  assert(1 <= x and x <= n);
  return x;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  int level = d(1);
  if (!level) {
    cout << "! 1" << endl;
    return 0;
  }
  int cur = 1;
  while (1) {
    int to = get(cur);
    int res = d(to);
    if (!res) {
      cout << "! " << to << endl;
      break;
    }
    if (lvl[to] + res != level) to = ana(to, (lvl[to] - level + res) / 2);
    if (lvl[to] == level) {
      cout << "! " << to << endl;
      break;
    }
    vis[to] = 1;
    cur = s(to);
  }
  assert(1 <= ask and ask <= 36);
  return 0;
}