description: Lord Omkar would like to have a tree with n nodes (3 ≤ n ≤ 10^5) and has asked his disciples to construct the tree. However, Lord Omkar has created m (1 ≤ m < n) restrictions to ensure that the tree will be as heavenly as possible. 
  
  A tree with n nodes is an connected undirected graph with n nodes and n-1 edges. Note that for any two nodes, there is exactly one simple path between them, where a simple path is a path between two nodes that does not contain any node more than once.
  
  Here is an example of a tree: 
  
  <image>
  
  A restriction consists of 3 pairwise distinct integers, a, b, and c (1 ≤ a,b,c ≤ n). It signifies that node b cannot lie on the simple path between node a and node c. 
  
  Can you help Lord Omkar and become his most trusted disciple? You will need to find heavenly trees for multiple sets of restrictions. It can be shown that a heavenly tree will always exist for any set of restrictions under the given constraints.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  The first line of each test case contains two integers, n and m (3 ≤ n ≤ 10^5, 1 ≤ m < n), representing the size of the tree and the number of restrictions.
  
  The i-th of the next m lines contains three integers a_i, b_i, c_i (1 ≤ a_i, b_i, c_i ≤ n, a, b, c are distinct), signifying that node b_i cannot lie on the simple path between nodes a_i and c_i. 
  
  It is guaranteed that the sum of n across all test cases will not exceed 10^5.
  
  Output
  
  For each test case, output n-1 lines representing the n-1 edges in the tree. On each line, output two integers u and v (1 ≤ u, v ≤ n, u ≠ v) signifying that there is an edge between nodes u and v. Given edges have to form a tree that satisfies Omkar's restrictions.
  
  Example
  
  Input
  
  
  2
  7 4
  1 2 3
  3 4 5
  5 6 7
  6 5 4
  5 3
  1 2 3
  2 3 4
  3 4 5
  
  
  Output
  
  
  1 2
  1 3
  3 5
  3 4
  2 7
  7 6
  5 1
  1 3
  3 2
  2 4
  
  Note
  
  The output of the first sample case corresponds to the following tree: 
  
  <image> For the first restriction, the simple path between 1 and 3 is 1, 3, which doesn't contain 2. The simple path between 3 and 5 is 3, 5, which doesn't contain 4. The simple path between 5 and 7 is 5, 3, 1, 2, 7, which doesn't contain 6. The simple path between 6 and 4 is 6, 7, 2, 1, 3, 4, which doesn't contain 5. Thus, this tree meets all of the restrictions.
  
  The output of the second sample case corresponds to the following tree: 
  
  <image>
solution: import java.util.*;
  import java.io.*;
  
  public class B_1586 {
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		
  		int t = sc.nextInt();
  		while(t-->0) {
  			int n = sc.nextInt(), m = sc.nextInt();
  			TreeSet<Integer> set = new TreeSet<>();
  			for(int i = 0; i < n; i++)
  				set.add(i);
  			for(int i = 0; i < m; i++) {
  				sc.nextInt();
  				set.remove(sc.nextInt() - 1);
  				sc.nextInt();
  			}
  			int root = set.first();
  			for(int i = 0; i < n; i++) {
  				if(i == root)
  					continue;
  				pw.println((root + 1) + " " + (i + 1));
  			}
  		}
  		
  		pw.flush();
  	}
  
  	public static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextIntArray(int n) throws IOException {
  			int[] array = new int[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextInt();
  			return array;
  		}
  
  		public Integer[] nextIntegerArray(int n) throws IOException {
  			Integer[] array = new Integer[n];
  			for (int i = 0; i < n; i++)
  				array[i] = new Integer(nextInt());
  			return array;
  		}
  
  		public long[] nextLongArray(int n) throws IOException {
  			long[] array = new long[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextLong();
  			return array;
  		}
  
  		public double[] nextDoubleArray(int n) throws IOException {
  			double[] array = new double[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextDouble();
  			return array;
  		}
  
  		public static int[] shuffle(int[] a) {
  			int n = a.length;
  			Random rand = new Random();
  			for (int i = 0; i < n; i++) {
  				int tmpIdx = rand.nextInt(n);
  				int tmp = a[i];
  				a[i] = a[tmpIdx];
  				a[tmpIdx] = tmp;
  			}
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }
  