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 T1, class T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <class T1, class T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
const int MAXN = 200228;
int n;
vector<int> g[MAXN];
bool can[MAXN];
int deep[MAXN];
int cnts[MAXN];
bool used[MAXN];
int dist[MAXN];
vector<int> v[MAXN];
bool useds[MAXN];
void dfs(int u, int pr = -1) {
  can[u] = true;
  for (auto h : g[u]) {
    if (h != pr) {
      v[u].push_back(h);
      deep[h] = deep[u] + 1;
      dfs(h, u);
    }
  }
}
void recalc(int u) {
  cnts[u] = can[u];
  for (auto h : v[u]) {
    recalc(h);
    cnts[u] += cnts[h];
  }
}
queue<int> q;
void makethempay(int u, int dists) {
  for (int i = 0; i < n; i++) {
    used[i] = false;
    dist[i] = 0;
  }
  q.push(u);
  used[u] = true;
  while (!q.empty()) {
    auto a = q.front();
    q.pop();
    can[a] &= (dist[a] == dists);
    for (auto h : g[a]) {
      if (!used[h]) {
        used[h] = true;
        dist[h] = dist[a] + 1;
        q.push(h);
      }
    }
  }
  recalc(0);
}
int getdist(int u) {
  cout << 'd' << ' ' << u + 1 << endl;
  int res;
  cin >> res;
  if (res == -1) {
    exit(0);
  }
  return res;
}
mt19937 as;
vector<int> st;
void get(int u) {
  st.push_back(u);
  for (auto h : v[u]) {
    get(h);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  as.seed(time(NULL));
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  useds[0] = true;
  int u = 0;
  int need = getdist(u);
  makethempay(u, need);
  int level = 0;
  while (true) {
    if (level == need) {
      cout << '!' << ' ' << u + 1 << endl;
      return 0;
    }
    while (true) {
      int kek = 0;
      int ress = 0;
      int ft = 0;
      int pos = -1;
      int who;
      for (auto h : v[u]) {
        if (cnts[h]) {
          chkmax(kek, cnts[h]);
          ress += cnts[h];
          ft++;
          pos = h;
          if (kek == cnts[h]) {
            who = h;
          }
        }
      }
      assert(ft != 0);
      if (ft == 1) {
        u = pos;
        level++;
        break;
      }
      if (kek * 2 <= ress) {
        cout << 's' << ' ' << u + 1 << endl;
        int res;
        cin >> res;
        if (res == -1) {
          exit(0);
        }
        res--;
        u = res;
        level++;
        break;
      }
      st.clear();
      get(who);
      int f = st[as() % (int)(st).size()];
      while (useds[f]) {
        f = st[as() % (int)(st).size()];
      }
      useds[f] = true;
      makethempay(f, getdist(f));
    }
  }
  return 0;
}