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;
vector<int> g[200010];
const int logn = 19;
int dp[20][200010];
int sub[200010];
int dep[200010];
int nxt[200010];
int tail[200010];
int n;
int distance(int x) {
  cout << "d " << x << endl;
  int ans;
  cin >> ans;
  return ans;
}
int go(int x) {
  cout << "s " << x << endl;
  int ans;
  cin >> ans;
  return ans;
}
int lift(int x, int up) {
  for (int i = logn; i >= 0; i--) {
    if (up >= (1 << i)) {
      up -= 1 << i;
      x = dp[i][x];
    }
  }
  return x;
}
void dfs(int x, int par) {
  sub[x] = 1;
  nxt[x] = -1;
  dp[0][x] = par;
  for (auto i : g[x]) {
    if (i - par) {
      dep[i] = 1 + dep[x];
      dfs(i, x);
      sub[x] += sub[i];
      if (nxt[x] == -1 || sub[nxt[x]] < sub[i]) {
        nxt[x] = i;
      }
    }
  }
}
void build() {
  for (int i = 1; i <= n; i++) {
    if (i == 1 || nxt[dp[0][i]] != i) {
      tail[i] = i;
      for (int j = i; j != -1; j = nxt[j]) {
        tail[i] = j;
      }
    }
  }
  for (int i = 1; i <= logn; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = dp[i - 1][dp[i - 1][j]];
    }
  }
}
int solver(int x, int dist) {
  if (dist == 0) return x;
  int y = tail[x];
  int ydist = distance(y);
  int p = ydist - dist;
  int q = dep[y] - dep[x];
  int up = (p + q) / 2;
  int node = lift(y, up);
  int dist_node = dist - (dep[node] - dep[x]);
  if (dist_node == 0) return node;
  return solver(go(node), dist_node - 1);
}
int main(int argc, char const *argv[]) {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p, q;
    cin >> p >> q;
    g[p].push_back(q);
    g[q].push_back(p);
  }
  dfs(1, 0);
  build();
  int x = solver(1, distance(1));
  cout << "! " << x << endl;
  return 0;
}