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;
  const int MAX = 1e5 + 1;
  const int MOD = 1e9 + 7;
  const double EPS = 1e-8;
  set<pair<int, string>> S;
  void search(string cur, int i, int j) {
    S.insert({cur.size(), cur});
    if (cur.size() >= 10) return;
    if (cur.size() || i != 0) search(cur + char(i + '0'), i, j);
    if (cur.size() != 0 || j != 0) search(cur + char(j + '0'), i, j);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    for (int i = 0; i <= 9; ++i)
      for (int j = i; j <= 9; ++j) search("", i, j);
    int tt;
    cin >> tt;
    while (tt--) {
      string n, temp = "";
      int k;
      cin >> n >> k;
      if (k == 1) {
        for (int i = 1; i <= 9; ++i) {
          for (int j = 0; j < n.size(); ++j) {
            temp += char(i + '0');
          }
          if (temp >= n) {
            cout << temp;
            break;
          }
          temp = "";
        }
      } else {
        cout << S.lower_bound({n.size(), n})->second;
      }
      cout << '\n';
    }
  }
  