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 get_ans(long long int i, long long int n, long long int distinct,
                 string nw, long long int k, string str) {
    if (distinct > k) {
      return "";
    }
    if (nw.substr(0, i) < str.substr(0, i)) {
      return "";
    }
    if (i >= n) return nw;
    for (char ch = '0'; ch <= '9'; ch++) {
      nw[i] = ch;
      bool b = true;
      for (long long int j = 0; j < i; j++) {
        if (nw[j] == ch) {
          b = false;
          break;
        }
      }
      string temp = get_ans(i + 1, n, distinct + b, nw, k, str);
      if (temp.size()) {
        return temp;
      }
    }
    return "";
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      string str = to_string(n);
      string ans = get_ans(0, str.size(), 0, string(str.size(), '0'), k, str);
      cout << ans << "\n";
    }
  }
  