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;
  vector<int> v;
  const char nl = '\n';
  string solve1(string s, int n) {
    string t;
    for (char c = '1'; c <= '9'; c++) {
      t = string(n, c);
      if (t >= s) break;
    }
    return t;
  }
  string solve2(string s, int n) {
    string res = solve1(s, n);
    for (char a = '0'; a < '9'; a++) {
      for (char b = a + 1; b <= '9'; b++) {
        bool only_ab = true;
        for (int i = 0; i < n; i++) {
          if (s[i] < b) {
            string t = s;
            if (s[i] < a)
              t[i] = a;
            else
              t[i] = b;
            for (int k = i + 1; k < n; k++) t[k] = a;
            if (t < res) res = t;
          }
          if (s[i] != a && s[i] != b) {
            only_ab = false;
            break;
          }
        }
        if (only_ab) return s;
      }
    }
    return res;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k, t;
    cin >> t;
    while (t--) {
      string s;
      int k;
      cin >> s >> k;
      int n = s.length();
      k == 1 ? cout << solve1(s, n) : cout << solve2(s, n);
      cout << nl;
    }
  }
  