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;
  const int maxn = 2 * 1e5 + 10, maxm = 1e6 + 10, mod = 998244353;
  const double pi = acos(-1), eps = 1e-7;
  int T;
  int n, k;
  set<char> se;
  int cnt(string s) {
    set<char> ses;
    for (char i : s) {
      ses.insert(i);
    }
    return ses.size();
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while (T--) {
      cin >> n >> k;
      string s = to_string(n);
      se.clear();
      int m = s.size();
      if (cnt(s) <= k) {
        cout << n << "\n";
      } else {
        if (k == 1) {
          for (char i = s[0]; i <= '9'; i++) {
            string str = "";
            for (int j = 0; j < m; j++) {
              str += i;
            }
            if (str >= s) {
              cout << str << "\n";
              break;
            }
          }
        } else {
          string ans = "9999999999";
          for (int i = (m - 1); i >= 0; i--) {
            se.clear();
            string tem = "";
            for (int j = 0; j < i; j++) {
              se.insert(s[j]);
              tem += s[j];
            }
            int t = se.size();
            if (t > k) continue;
            for (char j = s[i] + 1, cur; j <= '9'; j++) {
              if (se.count(j))
                cur = t;
              else
                cur = t + 1;
              if (cur > k) continue;
              char minv;
              if (cur + 1 <= k)
                minv = '0';
              else
                minv = min(j, *se.begin());
              string res = tem;
              res += j;
              for (int k = i + 1; k < m; k++) {
                res += minv;
              }
              ans = min(ans, res);
            }
          }
          cout << ans << "\n";
        }
      }
    }
    return 0;
  }
  