description: It is a simplified version of problem F2. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  4
  1 1
  221 2
  177890 2
  998244353 1
  
  
  Output
  
  
  1
  221
  181111
  999999999
solution: // package Codeforce.cf739;
  
  import java.io.PrintWriter;
  import java.util.Scanner;
  import java.util.TreeSet;
  
  public class F {
      //    MUST SEE BEFORE SUBMISSION
  //    check whether int part would overflow or not, especially when it is a * b!!!!
  //    check if top down dp would cause overflow or not !!!!!!!!!!!!!!!!!!!!!!
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          PreCompute();
  //        int t = 1;
          for (int i = 0; i < t; i++) {
              solve(sc, pw);
          }
          pw.close();
      }
      static TreeSet<Long> set = new TreeSet<>();
      static void PreCompute(){
          for(int i = 0; i < 10; i++){
              for(int j = i + 1; j < 10; j++){
                  for(int l = 2; l <= 10; l++){
                      for(int t = 1; t < (1 << l) - 1; t++){
                          int[] neo = new int[l];
                          for (int k = 0; k < l; k++) {
                              if ((t | (1 << k)) == t){
                                  neo[k] = i;
                              }else{
                                  neo[k] = j;
                              }
                          }
                          if (neo[0] == 0) continue;
                          long v = 0;
                          for (int k = 0; k < l; k++) {
                              v *= 10;
                              v += neo[k];
                          }
                          set.add(v);
                      }
                  }
              }
          }
      }
  
      static void solve(Scanner in, PrintWriter out){
          int n = in.nextInt(), k = in.nextInt();
          int len = String.valueOf(n).length();
          if (k == 1){
              long mut = 0;
              for(int i = 0; i < len; i++){
                  mut *= 10;
                  mut += 1;
              }
              for(int i = 1; i < 10; i++){
                  if (i * mut >= n){
                      out.println(i * mut);
                      return;
                  }
              }
          }else{
              long mut = 0;
              long min = 0;
              for(int i = 0; i < len; i++){
                  mut *= 10;
                  mut += 1;
              }
              for(int i = 1; i < 10; i++){
                  if (i * mut >= n){
                      min = (i * mut);
                      break;
                  }
              }
              out.println(Math.min(set.ceiling((long) n), min));
          }
      }
  }
  