D. Kuroni and the Celebration

This is an interactive problem.

After getting AC after 13 Time Limit Exceeded verdicts on a geometry problem, Kuroni went to an Italian restaurant to celebrate this holy achievement. Unfortunately, the excess sauce disoriented him, and he's now lost!

The United States of America can be modeled as a tree (why though) with nn vertices. The tree is rooted at vertex rr, wherein lies Kuroni's hotel.

Kuroni has a phone app designed to help him in such emergency cases. To use the app, he has to input two vertices uu and vv, and it'll return a vertex ww, which is the lowest common ancestor of those two vertices.

However, since the phone's battery has been almost drained out from live-streaming Kuroni's celebration party, he could only use the app at most ⌊n2⌋⌊n2⌋ times. After that, the phone would die and there will be nothing left to help our dear friend! :(

As the night is cold and dark, Kuroni needs to get back, so that he can reunite with his comfy bed and pillow(s). Can you help him figure out his hotel's location?

Interaction

The interaction starts with reading a single integer nn (2≤n≤10002≤n≤1000), the number of vertices of the tree.

Then you will read n−1n−1 lines, the ii-th of them has two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi), denoting there is an edge connecting vertices xixi and yiyi. It is guaranteed that the edges will form a tree.

Then you can make queries of type "? u v" (1≤u,v≤n1≤u,v≤n) to find the lowest common ancestor of vertex uu and vv.

After the query, read the result ww as an integer.

In case your query is invalid or you asked more than ⌊n2⌋⌊n2⌋ queries, the program will print −1−1 and will finish interaction. You will receive a Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.

When you find out the vertex rr, print "! rr" and quit after that. This query does not count towards the ⌊n2⌋⌊n2⌋ limit.

Note that the tree is fixed beforehand and will not change during the queries, i.e. the interactor is not adaptive.

After printing any query do not forget to print end of line and flush the output. Otherwise, you might 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

To hack, use the following format:

The first line should contain two integers nn and rr (2≤n≤10002≤n≤1000, 1≤r≤n1≤r≤n), denoting the number of vertices and the vertex with Kuroni's hotel.

The ii-th of the next n−1n−1 lines should contain two integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n) — denoting there is an edge connecting vertex xixi and yiyi.

The edges presented should form a tree.

Example

input

Copy

6
1 4
4 2
5 3
6 3
2 3

3

4

4

output

Copy

? 5 6

? 3 1

? 1 2

! 4

Note

Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.

The image below demonstrates the tree in the sample test:

 

Code：队列维护叶子节点

#include <bits/stdc++.h>
using namespace std;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define pb push_back
typedef pair<int,int> pii;
const ll mod = 1000000000 ;
const int INF=0x3f3f3f3f;
using namespace std;
const int mxn = 3e5+7 ;
int n,t,m,k,l,r,res,prime[mxn],isprime[mxn],phi[mxn],dis[mxn],head[mxn];
string str;
vector<int>ans[mxn];
queue<int>q;
int cnt[mxn];
bool vis[mxn];
void del(int x)
{
    for(auto i : ans[x])
    {
        cnt[i]--;
        if(cnt[i]==1) q.push(i);
    }
}
int main()
{
    TLE;
    scanf("%d",&n);
    for(int i=1,u,v;i<n;i++)
    {
        scanf("%d %d",&u,&v);
        ans[u].pb(v);ans[v].pb(u);
        cnt[u]++;cnt[v]++;
    }
    for(int i=1;i<=n;i++)
        if(cnt[i]==1) q.push(i);
    while(q.size()>=2)
    {
        int u = q.front() ;q.pop();
        int v = q.front() ;q.pop();
        vis[u] = vis[v] = 1 ;
        printf("? %d %d\n",u,v);
        fflush(stdout);
        cin>>k;
        if(k==u || k==v ) {printf("! %d\n",k); return 0;}
        del(u);del(v);
    }
    k=0;
    for(int i=1;i<=n;i++)
        if(!vis[i])
            printf("! %d\n",i) , k++;
    assert(k==1);
    return 0 ;
}