Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T. 
For example, consider the tree: 
<center> </center>
Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two. 

For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number. 

Input

The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.

Output

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

Sample Input

1
7
2 6
1 2
1 4
4 5
3 7
3 1

Sample Output

1 2


题意就是给你一棵无根树，让你找到一个点，去掉这个点之后使得剩下的子树的最大节点数最小；

思路，就是求树的重心，下面先给出树的重心的定义：对于一棵n个节点的无根树，找到一个点使得把树变成一棵以该节点为根的有根树，这时的最大子树的节点数最小。
定义sizes[i]表示i的最大子树的节点数，定义dp[i]为以i为根的最大子树的节点数。然后递归求解。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdio>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <cstring>
 7 using namespace std;
 8 typedef long long ll;
 9 const int maxn=1e6+5; 11 int dp[maxn];
12 int sizes[maxn];
13 int ans,n,sum;
14 vector<int> v[maxn];//二维矩阵存图
15 void dfs(int x,int fa)
16 {
17     sizes[x]=1;
18     int maxx=0;
19     for(int i=0;i<v[x].size();i++)
20     {
21         int y=v[x][i];
22         if(y!=fa)
23         {
24             dfs(y,x);
25             sizes[x]+=sizes[y];
26             maxx=max(maxx,sizes[y]);
27         }
28 
29     }
30         dp[x]=max(maxx,n-sizes[x]);
31         if(sum>dp[x])
32         {
33             ans=x;
34             sum=dp[x];
35         }
36 }
37 int main()
38 {
39     int T;
40     scanf("%d",&T);
41     while(T--)
42     {
43         sum=0x3f3f3f3f;
44         scanf("%d",&n);
45         memset(dp,0,sizeof(dp));
46         memset(sizes,0,sizeof(sizes));
47         for(int i=1;i<=n;i++)v[i].clear();
48         for(int i=1;i<n;i++)
49         {
50             int L,K;
51             scanf("%d%d",&L,&K);
52             v[K].push_back(L);
53             v[L].push_back(K);
54         }
55         dfs(1,-1);
56         printf("%d %d\n",ans,dp[ans]);
57     }
58 
59 
60     return 0;
61 }