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;
  string solve1(string s) {
    string temp1 = "";
    string temp2 = "";
    if (s[0] != 9) {
      for (int i = 0; i < s.length(); i++) {
        temp1 += s[0];
        temp2 += s[0] + 1;
      }
      if (temp1 >= s) {
        return temp1;
      } else {
        return temp2;
      }
    } else {
      for (int i = 0; i < s.length(); i++) {
        temp1 += s[0];
      }
      return temp1;
    }
  }
  void solve2(string s) {
    string res = solve1(s);
    for (char a = '0'; a <= '9'; a++) {
      for (char b = '1'; b <= '9'; b++) {
        bool s_ok = true;
        for (int i = 0; i < s.length(); i++) {
          if (s[i] < b) {
            string t = s;
            if (t[i] < a) {
              t[i] = a;
            } else {
              t[i] = b;
            }
            for (int j = i + 1; j < s.length(); j++) {
              t[j] = a;
            }
            if (res > t) {
              res = t;
            }
          }
          if (s[i] != a && s[i] != b) {
            s_ok = false;
            break;
          }
        }
        if (s_ok) {
          cout << s << endl;
          return;
        }
      }
    }
    cout << res << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      string s = "";
      long long int temp = n;
      while (temp != 0) {
        s.push_back((temp % 10) + '0');
        temp /= 10;
      }
      reverse(s.begin(), s.end());
      if (k == 1) {
        cout << solve1(s) << endl;
      } else {
        solve2(s);
      }
    }
    return 0;
  }
  