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;
  using namespace std;
  const int N = 500005;
  const long long Mod = 1e9 + 7;
  int main() {
    int t, k;
    string n;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      string ans;
      for (char a = '0'; a <= '9'; a++) {
        string t(n.length(), a);
        if (t >= n) {
          ans = t;
          break;
        }
      }
      if (k == 1) {
        cout << ans << endl;
      } else {
        for (char a = '0'; a <= '9'; a++) {
          for (char b = a + 1; b <= '9'; b++) {
            bool ok = true;
            for (int i = 0; i < n.length(); i++) {
              if (n[i] < b) {
                string t = n;
                if (n[i] < a)
                  t[i] = a;
                else
                  t[i] = b;
                for (int j = i + 1; j < n.length(); j++) {
                  t[j] = a;
                }
                if (ans > t) ans = t;
              }
              if (n[i] != a && n[i] != b) {
                ok = false;
                break;
              }
            }
            if (ok) ans = n;
          }
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  