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 T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const long long &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1e6 + 10;
int lv[SIZE], lat[SIZE];
vector<int> e[SIZE];
int can[SIZE];
int AA;
void dfs0(int x, int lt, int v) {
  lv[x] = v;
  for (int y : e[x]) {
    if (y == lt) continue;
    lat[y] = x;
    dfs0(y, x, v + 1);
  }
}
int target;
int cnt(int x, int lt, int dd) {
  if (!dd) {
    target = x;
    return 1;
  }
  int ret = 0;
  for (int y : e[x]) {
    if (y == lt) continue;
    ret += cnt(y, x, dd - 1);
  }
  return ret;
}
int dp[SIZE];
int dfs2(int x, int lt, int dd) {
  if (!dd) {
    dp[x] = can[x];
    return -1;
  }
  dp[x] = 0;
  for (int y : e[x]) {
    if (y == lt) continue;
    int now = dfs2(y, x, dd - 1);
    if (now != -1) return now;
    dp[x] += dp[y];
    if (dp[x] >= (AA + 1) / 2) return x;
  }
  return -1;
}
void clear(int x, int lt, int dd) {
  if (!dd) {
    if (can[x]) {
      can[x] = 0;
      AA--;
    }
    return;
  }
  for (int y : e[x]) {
    if (y == lt) continue;
    clear(y, x, dd - 1);
  }
}
void clear2(int x, int lt, int stop, int dd) {
  if (x == stop) return;
  if (!dd) {
    if (can[x]) {
      can[x] = 0;
      AA--;
    }
    return;
  }
  for (int y : e[x]) {
    if (y == lt) continue;
    clear2(y, x, stop, dd - 1);
  }
}
int find_answer() {
  for (int i = (1); i <= (SIZE); ++i) {
    if (can[i]) return i;
  }
  return -1;
}
int qq(int x, int dd) {
  if (AA == 1) {
    return find_answer();
  }
  if (!dd) return x;
  int qx = dfs2(x, lat[x], dd);
  if (qx != x) {
    W("d", qx);
    fflush(stdout);
    int v;
    R(v);
    if (lv[qx] + v != lv[x] + dd) {
      clear(qx, lat[qx], lv[x] + dd - lv[qx]);
      qx = x;
    } else {
      clear2(x, lat[x], qx, dd);
    }
  }
  if (AA == 1) {
    return find_answer();
  }
  W("s", qx);
  fflush(stdout);
  int v;
  R(v);
  clear2(x, lat[x], v, dd);
  return qq(v, dd - (lv[v] - lv[x]));
}
int main() {
  int n;
  R(n);
  for (int i = (1); i < (n); ++i) {
    int x, y;
    R(x, y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs0(1, 1, 0);
  int now = 1, dd;
  W("d", now);
  fflush(stdout);
  R(dd);
  for (int i = (1); i <= (n); ++i) {
    if (lv[i] == dd) {
      can[i] = 1;
      AA++;
    }
  }
  W("!", qq(1, dd));
  return 0;
}