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 long long maxn = 200000;
bool used[maxn];
vector<long long> a[maxn];
vector<long long> b[maxn];
long long sub[maxn];
long long corn[maxn];
bool use[maxn];
long long pr[maxn];
void dfs(long long x) {
  used[x] = true;
  long long v;
  long long sz = 1;
  for (long long i = 0; i < a[x].size(); ++i) {
    v = a[x][i];
    pr[v] = x;
    if (!used[v] && !use[b[x][i]]) {
      corn[v] = corn[x] + 1;
      dfs(v);
      sz += sub[v];
    }
  }
  sub[x] = sz;
  used[x] = false;
}
long long go(long long x, long long sz) {
  used[x] = true;
  long long v;
  bool h = true;
  vector<long long> z;
  h = (sub[x] >= (sz / 2));
  long long ans = (-1);
  for (long long i = 0; i < a[x].size(); ++i) {
    v = a[x][i];
    if (!used[v] && !use[b[x][i]]) {
      ans = max(ans, go(v, sz));
      h = (h && sub[v] < (sz / 2));
    }
  }
  used[x] = false;
  if (h) {
    return x;
  } else {
    return ans;
  }
}
long long centroid(long long x) {
  dfs(x);
  return go(x, sub[x]);
}
void getans(long long x) {
  cout << "! " << (x + 1) << endl;
  exit(0);
}
long long dquery(long long x) {
  cout << "d " << (x + 1) << endl;
  long long ans;
  cin >> ans;
  if (ans == 0) {
    getans(x);
  }
  return ans;
}
long long squery(long long x) {
  cout << "s " << (x + 1) << endl;
  long long ans;
  cin >> ans;
  if (ans == (-1)) {
    exit(0);
  } else {
    return (ans - 1);
  }
}
long long sl(long long x, long long l) {
  corn[x] = 0;
  dfs(x);
  if (sub[x] == 1) {
    getans(x);
  }
  long long t = centroid(x);
  long long l2 = dquery(t);
  long long f = corn[t];
  if ((f + l2) == l) {
    long long o = squery(t);
    for (long long i = 0; i < a[t].size(); ++i) {
      use[b[t][i]] = true;
    }
    sl(o, l2 - 1);
  } else {
    long long o = pr[t];
    for (long long i = 0; i < a[t].size(); ++i) {
      use[b[t][i]] = true;
    }
    sl(x, l);
  }
}
int32_t main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (n - 1); ++i) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
    b[x].push_back(i);
    b[y].push_back(i);
  }
  sl(0, dquery(0));
  return 0;
}