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 maxn = 2e5 + 7;
const int INF = 0x3f3f3f3f;
vector<int> G[maxn];
int askdis(int x) {
  cout << 'd' << " " << x << endl;
  int ret;
  cin >> ret;
  return ret;
}
int askson(int x) {
  cout << 's' << " " << x << endl;
  int ret;
  cin >> ret;
  return ret;
}
void answer(int x) { cout << "! " << x << endl; }
int sz[maxn], maxson[maxn], rt, treesz;
int dep[maxn], far[maxn];
bool vis[maxn];
void dfs(int u, int fa = -1) {
  far[u] = fa;
  for (auto& v : G[u])
    if (v != fa) {
      dep[v] = dep[u] + 1;
      dfs(v, u);
    }
}
void dfsrt(int u, int fa = -1) {
  sz[u] = 1;
  maxson[u] = 0;
  for (auto& v : G[u])
    if (v != fa && !vis[v]) {
      dfsrt(v, u);
      sz[u] += sz[v];
      maxson[u] = max(maxson[u], sz[v]);
    }
  maxson[u] = max(maxson[u], treesz - sz[u]);
  if (maxson[rt] > maxson[u]) rt = u;
}
int getsz(int u, int fa = -1) {
  int ret = 1;
  for (auto& v : G[u])
    if (fa != v && !vis[v]) {
      ret += getsz(v, u);
    }
  return ret;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dep[1] = 0;
  dfs(1);
  treesz = n;
  int dis = askdis(1);
  int cur = 1;
  while (treesz != 1 && dis) {
    rt = 0;
    maxson[rt] = INF;
    dfsrt(cur);
    vis[rt] = 1;
    int dd = askdis(rt);
    if (!dd) {
      cur = rt;
      break;
    }
    if (dd + dep[rt] == dis) {
      cur = askson(rt);
      treesz = getsz(cur);
    } else {
      cur = far[rt];
      treesz = getsz(cur);
    };
  }
  answer(cur);
}