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;
  bool works(string s, char c) {
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] > c) return false;
      if (s[i] < c) return true;
    }
    return true;
  }
  void recur(string &s, string &res, string &cur, char a, char b) {
    if (cur.size() == s.size()) {
      if (cur[0] == 0) return;
      if (cur < s) return;
      if (res.size() == 0)
        res = cur;
      else if (cur < res)
        res = cur;
      return;
    }
    cur += a;
    recur(s, res, cur, a, b);
    cur.pop_back();
    cur += b;
    recur(s, res, cur, a, b);
    cur.pop_back();
  }
  string getComb(string s, string res, char a, char b) {
    string cur = "";
    recur(s, res, cur, a, b);
    if (b == '9') return res;
    b++;
    recur(s, res, cur, a, b);
    return res;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      string s = to_string(n);
      if (k == 1) {
        if (works(s, s[0]))
          cout << string(s.size(), s[0]);
        else
          cout << string(s.size(), s[0] + 1);
      } else {
        char a = s[0];
        char b = '*';
        for (long long i = 0; i < s.size(); i++) {
          if (s[i] != a) {
            b = s[i];
            break;
          }
        }
        if (b == '*')
          cout << string(s.size(), a);
        else {
          string res = string(s.size(), s[0] + 1);
          bool found = false;
          string x = "", y = "", z = "";
          for (long long i = 0; i < s.size(); i++) {
            if (s[i] != a) {
              if (s[i] < a) {
                x += a;
                x += string(s.size() - i - 1, '0');
              }
              recur(s, res, y, a, s[i]);
              if (s[i] != '9') recur(s, res, z, a, s[i] + 1);
              if (x.size() == s.size() && x >= s) res = min(res, x);
              if (y.size() == s.size() && y >= s) res = min(res, y);
              if (z.size() == s.size() && z >= s) res = min(res, z);
              break;
            } else
              x += s[i];
          }
          cout << res;
        }
      }
      cout << '\n';
    }
    return 0;
  }
  