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;
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long tt = 1;
    cin >> tt;
    vector<long long> k1;
    vector<long long> k2;
    for (long long d = 0; d <= 9; d++) {
      for (long long m = 1; m <= 10; m++) {
        long long z = 0;
        for (long long i = 0; i < m; i++) {
          z = z * 10 + d;
        }
        k1.push_back(z);
      }
    }
    for (long long m = 0; m <= 10; m++) {
      for (long long d1 = 0; d1 <= 9; d1++) {
        for (long long d2 = 0; d2 <= 9; d2++) {
          for (long long mask = 0; mask < (1 << m); mask++) {
            long long z = 0;
            for (long long i = 0; i < m; i++) {
              if (mask & (1 << i))
                z = z * 10 + d1;
              else
                z = z * 10 + d2;
            }
            k2.push_back(z);
          }
        }
      }
    }
    sort(k1.begin(), k1.end());
    sort(k2.begin(), k2.end());
    while (tt--) {
      long long n, k;
      cin >> n >> k;
      if (k == 1)
        cout << *lower_bound(k1.begin(), k1.end(), n) << '\n';
      else
        cout << *lower_bound(k2.begin(), k2.end(), n) << '\n';
    }
    return 0;
  }
  