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 int inf = 1e9;
const long long INF = 1e18;
const int M = 1e9 + 7;
const int maxn = 3e5;
int c, sz[maxn], st, n, d[maxn], fa[maxn], f[maxn];
std::vector<int> G[maxn];
int ok = 1, vis[maxn];
void dfs(int u, int p) {
  sz[u] = 1;
  int cur = 0;
  if (ok) {
    d[u] = d[p] + 1;
    fa[u] = p;
  }
  f[u] = p;
  for (int i : G[u]) {
    if (i == p || vis[i]) continue;
    dfs(i, u);
    cur = max(sz[i], cur);
    sz[u] += sz[i];
  }
  cur = max(n - sz[u], cur);
  if (cur < st) {
    c = u;
    st = cur;
  }
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  st = inf;
  d[0] = -1;
  dfs(1, 0);
  ok = 0;
  int dx;
  cout << "d " << 1 << endl;
  cin >> dx;
  while (1) {
    cout << "d " << c << endl;
    int t;
    cin >> t;
    if (t == 0) {
      cout << "! " << c << endl;
      break;
    }
    if (d[c] + t == dx) {
      cout << "s " << c << endl;
      cin >> t;
    } else {
      t = fa[c];
    }
    if (f[c] == t) {
      n -= sz[c];
    } else {
      n = sz[t];
    }
    st = inf;
    vis[c] = 1;
    dfs(t, 0);
  }
  return 0;
}