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 <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
template <typename T>
struct hash_map_t {
  vector<T> v, val, nxt;
  vector<int> head;
  int mod, tot, lastv;
  T lastans;
  bool have_ans;
  hash_map_t(int md = 0) {
    head.clear();
    v.clear();
    val.clear();
    nxt.clear();
    tot = 0;
    mod = md;
    nxt.resize(1);
    v.resize(1);
    val.resize(1);
    head.resize(mod);
    have_ans = 0;
  }
  void clear() { *this = hash_map_t(mod); }
  bool count(int x) {
    int u = x % mod;
    for (register int i = head[u]; i; i = nxt[i]) {
      if (v[i] == x) {
        have_ans = 1;
        lastv = x;
        lastans = val[i];
        return 1;
      }
    }
    return 0;
  }
  void ins(int x, int y) {
    int u = x % mod;
    nxt.push_back(head[u]);
    head[u] = ++tot;
    v.push_back(x);
    val.push_back(y);
  }
  int qry(int x) {
    if (have_ans && lastv == x) return lastans;
    count(x);
    return lastans;
  }
};
const int N = 2e5 + 5;
struct edge_t {
  int u, v, next;
} G[N << 1];
int head[N], fa[N], siz[N], wson[N], d[N], endpoint[N], top[N];
int n, tot, dep;
inline void addedge(int u, int v) {
  G[++tot] = (edge_t){u, v, head[u]}, head[u] = tot;
  G[++tot] = (edge_t){v, u, head[v]}, head[v] = tot;
}
inline int query1(int u) {
  cout << "d"
       << " " << u << endl;
  int ans;
  cin >> ans;
  return ans;
}
inline int query2(int u) {
  cout << "s"
       << " " << u << endl;
  int ans;
  cin >> ans;
  return ans;
}
void dfs1(int u, int f) {
  siz[u] = 1;
  for (register int i = head[u]; i; i = G[i].next) {
    int v = G[i].v;
    if (v == f) continue;
    d[v] = d[u] + 1;
    fa[v] = u;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[wson[u]]) wson[u] = v;
  }
}
void dfs2(int u, int tp) {
  top[u] = tp;
  if (!endpoint[tp])
    endpoint[tp] = u;
  else if (d[u] > d[endpoint[tp]])
    endpoint[tp] = u;
  if (wson[u]) dfs2(wson[u], tp);
  for (register int i = head[u]; i; i = G[i].next) {
    int v = G[i].v;
    if (v == fa[u] || v == wson[u]) continue;
    dfs2(v, v);
  }
}
int jump(int u, int cnt) {
  while (cnt) u = fa[u], --cnt;
  return u;
}
int ans = -1;
void query(int u) {
  int pt = endpoint[top[u]];
  int len = query1(pt);
  if (len == 0) {
    ans = pt;
    return;
  }
  if (d[pt] - len == dep) {
    ans = jump(pt, len);
    return;
  }
  int up = (len + d[pt] - dep) / 2;
  int det = (len - d[pt] + dep) / 2;
  pt = jump(pt, up);
  pt = query2(pt);
  if (det == 1) {
    ans = pt;
    return;
  }
  query(pt);
}
int main() {
  read(n);
  for (register int i = 1; i < n; i++) {
    int u, v;
    read(u);
    read(v);
    addedge(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  dep = query1(1);
  query(1);
  cout << "!"
       << " " << ans << endl;
  return 0;
}