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;
int tot = 0;
int qd(int x) {
  if (tot++ == 36)
    while (1)
      ;
  printf("d %d\n", x);
  fflush(stdout);
  int d;
  scanf("%d", &d);
  return d;
}
int qs(int x) {
  if (tot++ == 36)
    while (1)
      ;
  printf("s %d\n", x);
  fflush(stdout);
  int s;
  scanf("%d", &s);
  return s;
}
void ans(int x) {
  printf("! %d\n", x);
  fflush(stdout);
}
vector<int> edges[200005];
int sze[200005], dep[200005], par[200005];
int mxd;
void dfs(int u, int p, int d) {
  sze[u] = 0;
  par[u] = p;
  dep[u] = d;
  if (d == mxd) {
    sze[u] = 1;
    return;
  }
  for (auto v : edges[u]) {
    if (v == p) continue;
    dfs(v, u, d + 1);
    sze[u] += sze[v];
  }
}
int tmp[200005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  mxd = qd(1);
  if (!mxd) {
    ans(1);
    return 0;
  }
  dfs(1, -1, 0);
  int rt = 1;
  while (dep[rt] < mxd) {
    int s = sze[rt], hc = 0;
    for (auto u : edges[rt]) {
      if (u == par[rt]) continue;
      if (sze[u] * 2 > s) {
        hc = u;
        break;
      }
    }
    if (!hc) {
      rt = qs(rt);
    } else {
      int t = rt, it = 0;
      while (dep[t] < mxd) {
        bool large = 0;
        tmp[++it] = t;
        for (auto v : edges[t]) {
          if (v != par[t] && sze[v] * 2 > s) {
            t = v;
            large = 1;
            break;
          }
        }
        if (!large) break;
      }
      int d = qd(t);
      if (d == -1) return 0;
      rt = qs(tmp[it - (d - (mxd - dep[t])) / 2]);
    }
    if (rt == -1) return 0;
  }
  ans(rt);
  return 0;
}