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;
  set<int> ans1, ans2;
  int x1, x2;
  void rec(int i, int x) {
    ans1.insert(x);
    if (x1 == x2) {
      ans2.insert(x);
    }
    if (i == 9) {
      return;
    }
    rec(i + 1, x * 10 + x1);
    rec(i + 1, x * 10 + x2);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t, n, k;
    cin >> t;
    for (x1 = 0; x1 <= 9; x1++) {
      for (x2 = 0; x2 <= 9; x2++) {
        rec(0, 0);
      }
    }
    while (t--) {
      cin >> n >> k;
      if (n == 1e9) {
        if (k == 2)
          cout << n << "\n";
        else
          cout << "1111111111\n";
        continue;
      }
      if (k == 1) {
        cout << (*ans2.lower_bound(n)) << "\n";
      } else {
        cout << (*ans1.lower_bound(n)) << "\n";
      }
    }
  }
  