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 n) {
    string res(n.length(), '9');
    for (char c = '8'; c >= '0'; c--) {
      string t(n.length(), c);
      if (t >= n) res = t;
    }
    return res;
  }
  string solve2(string n) {
    string res = solve1(n);
    for (char a = '0'; a <= '9'; a++)
      for (char b = a + 1; b <= '9'; b++) {
        bool n_ok = true;
        for (int i = 0; i < n.length(); i++) {
          if (n[i] < b) {
            string t = n;
            if (t[i] < a)
              t[i] = a;
            else
              t[i] = b;
            for (int j = i + 1; j < n.length(); j++) t[j] = a;
            if (res > t) res = t;
          }
          if (n[i] != a && n[i] != b) {
            n_ok = false;
            break;
          }
        }
        if (n_ok) return n;
      }
    return res;
  }
  string solve() {
    string n;
    int k;
    cin >> n >> k;
    if (k == 1)
      return solve1(n);
    else
      return solve2(n);
  }
  int main() {
    int t;
    cin >> t;
    while (t--) cout << solve() << '\n';
    return 0;
  }
  