Problem: Fishing Prince loves trees, and he especially loves trees with only one centroid. The tree is a connected graph without cycles.

A vertex is a centroid of a tree only when you cut this vertex (remove it and remove all edges from this vertex), the size of the largest connected component of the remaining graph is the smallest possible.

For example, the centroid of the following tree is 2, because when you cut it, the size of the largest connected component of the remaining graph is 2 and it can't be smaller.

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However, in some trees, there might be more than one centroid, for example:

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Both vertex 1 and vertex 2 are centroids because the size of the largest connected component is 3 after cutting each of them.

Now Fishing Prince has a tree. He should cut one edge of the tree (it means to remove the edge). After that, he should add one edge. The resulting graph after these two operations should be a tree. He can add the edge that he cut.

He wants the centroid of the resulting tree to be unique. Help him and find any possible way to make the operations. It can be proved, that at least one such way always exists.

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (3≤ n≤ 10^5) — the number of vertices.

Each of the next n-1 lines contains two integers x, y (1≤ x,y≤ n). It means, that there exists an edge connecting vertices x and y.

It's guaranteed that the given graph is a tree.

It's guaranteed that the sum of n for all test cases does not exceed 10^5.

Output

For each test case, print two lines.

In the first line print two integers x_1, y_1 (1 ≤ x_1, y_1 ≤ n), which means you cut the edge between vertices x_1 and y_1. There should exist edge connecting vertices x_1 and y_1.

In the second line print two integers x_2, y_2 (1 ≤ x_2, y_2 ≤ n), which means you add the edge between vertices x_2 and y_2.

The graph after these two operations should be a tree.

If there are multiple solutions you can print any.

Example

Input


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


Output


1 2
1 2
1 3
2 3

Note

Note that you can add the same edge that you cut.

In the first test case, after cutting and adding the same edge, the vertex 2 is still the only centroid.

In the second test case, the vertex 2 becomes the only centroid after cutting the edge between vertices 1 and 3 and adding the edge between vertices 2 and 3.

Solution (in JAVA):

Solution:
import java.io.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
  //   static long mod=1000000007;
     static HashSet<Integer> adj[];
   //  static boolean vis[];
  //   static int n[];
  //   static ArrayList<Integer> arr[];
  // //  static long ans=0;
  //   static boolean vis[][];
  //   static final long oo=(long)1e18;
    // static int par[];
    // static long child[];
    // static HashMap<String,Long> hm=new HashMap<>();
        static int size[];
     public static void main(String[] args) throws IOException {  
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        int test=nextInt();
       // int test=1;
        outer: while(test--!=0){
            int n=nextInt();
            adj=new HashSet[n];
            for(int i=0;i<n;i++){
                adj[i]=new HashSet<>();
            }
            size=new int[n];
            int tempa=0,tempb=0;
            for(int i=0;i<n-1;i++){
                int a=nextInt()-1;
                int b=nextInt()-1;
                tempa=a+1;
                tempb=b+1;
                adj[a].add(b);
                adj[b].add(a);
            }
            int sz=dfs(0,-1);
         //   pw.println(sz);
            int a=-1;int b=-1;
            for(int i=0;i<n;i++){
                if(size[i]==n/2){
                    
                        a=i;
                    
                        
                }
            }
          //  pw.println(a+" "+b+" "+size[0]);
            if((a==-1&&b==-1)||n%2!=0){
                pw.println(tempa+" "+tempb);
                pw.println(tempa+" "+tempb);
                continue outer;
            }
            
            int v=-1;
            for(int ver:adj[a]){
                if(size[ver]>n/2)
                    b=ver;
                else
                    v=ver;
                
            }
           // pw.println(a+" "+b+" ");
            
                pw.println((a+1)+" "+(v+1));
                pw.println((b+1)+" "+(v+1));
                
            //}
            // else{
            //     for(int ver:adj[b]){
            //         if(ver!=a){
            //             v=ver;
            //         }
            //     }
            //     pw.println(b+" "+v);
            //     pw.println(a+" "+v);
            // }

        }
           
        pw.close();
    }
    static int dfs(int v,int par){
        int sz=1;
        for(int a:adj[v]){
            if(a!=par){
                sz+=dfs(a,v);
            }
        }
        size[v]=sz;
        return sz;
    }
    
    
    static class compare implements Comparable<compare>{
         String s;
         compare(String s){
            this.s=s;
        }
        public  int compareTo(compare t){
            return compareVersion(s,t.s);
        }
        public  int compareVersion(String version1, String version2) {
            String[] nums1 = version1.split("\\.");
            String[] nums2 = version2.split("\\.");
            int n1 = nums1.length, n2 = nums2.length;

            // compare versions
            int i1, i2;
            for (int i = 0; i < Math.max(n1, n2); ++i) {
              i1 = i < n1 ? Integer.parseInt(nums1[i]) : 0;
              i2 = i < n2 ? Integer.parseInt(nums2[i]) : 0;
              if (i1 != i2) {
                return i1 > i2 ? 1 : -1;
              }
            }
            // the versions are equal
            return 0;
        }
    }
  
    
    static class Pair{
        int a;int b;
        Pair(int a,int b){
          //  this.index=index;
            this.a=a;
            this.b=b;
            
            //this.z=z;
        }   
    }
   static void update(long tree[],int idx,long val){
        //println(idx);
        while(idx<1000006){
            
            tree[idx]+=val;
            idx+=(idx)&(-idx);
        }
    }   
    static long sum(long tree[],int idx){
        long sum=0;
        while(idx>0){
           
            sum+=tree[idx];
            idx-=(idx)&(-idx);
        }
        return sum;
    }
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    // static class Pair{
    //     int x;int y;
    //     Pair(int x,int y,int z){
    //         this.x=x;
    //         this.y=y;
    //        // this.z=z;
    //      //   this.z=z;
    //       //  this.i=i;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public int compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         if(a.y==b.y){
    //             return -1*(a.z-b.z);
    //         }
    //         return (a.y-b.y);
    //     }
   // }

    static long ncr(long n,long r){
        
        if(r==0)
            return 1l;
        long val=ncr(n-1,r-1);
        val=(n*val);
        val=(val/r);
        return val;
    }

    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    
    // static class sorting implements Comparator<Pair>{  
    //     public int compare(Pair a1,Pair a2){  
    //         if(o1.a==o2.a)
    //             return (o1.b>o2.b)?1:-1;  
    //         else if(o1.a>o2.a)  
    //             return 1;  
    //         else  
    //             return -1;  
    //     }
    // }  
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,long M){
        return fast_pow(n,M-2,M);
    }
    // (1,1)   

    //     (3,2)   (3,5)

}