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: #include <bits/stdc++.h>
  using namespace std;
  long long a, b;
  set<long long> s, y;
  void func(int l, long long num) {
    if (l <= 0) {
      s.insert(num);
      return;
    }
    func(l - 1, num * 10 + a);
    func(l - 1, num * 10 + b);
  }
  void fuck(int l, long long num) {
    if (l <= 0) {
      y.insert(num);
      return;
    }
    fuck(l - 1, num * 10 + a);
  }
  int main() {
    for (int i = 1; i <= 9; i++) s.insert(i);
    for (int i = 2; i <= 11; i++) {
      for (int j = 0; j <= 9; j++) {
        for (int k = 0; k <= 9; k++) {
          a = j, b = k;
          func(i, 0);
        }
      }
    }
    for (int i = 1; i <= 11; i++)
      for (int j = 0; j <= 9; j++) {
        a = j;
        fuck(i, 0);
      }
    int t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      if (k == 2)
        cout << *(s.lower_bound(n)) << endl;
      else
        cout << *(y.lower_bound(n)) << endl;
    }
    return 0;
  }
  