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 INF = 1e9 + 7;
  set<long long> s1;
  set<long long> s2;
  long long p[20];
  void gen() {
    p[0] = 1;
    for (long long i = 1; i <= 15; i++) p[i] = p[i - 1] * 10;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    if (k == 1)
      cout << *s1.lower_bound(n) << "\n";
    else
      cout << *s2.lower_bound(n) << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    gen();
    for (long long i = 1; i <= 9; i++) {
      long long liczba = 0;
      for (long long j = 1; j <= 10; j++) {
        liczba *= 10;
        liczba += i;
        s1.insert(liczba);
      }
    }
    for (long long i = 0; i <= 9; i++)
      for (long long j = i + 1; j <= 9; j++)
        for (long long l = 0; l <= 10; l++)
          for (long long mask = 0; mask <= (1 << l) - 1; mask++) {
            long long liczba = 0;
            for (long long m = 0; m <= l - 1; m++) {
              if (mask & (1 << m))
                liczba += p[m] * i;
              else
                liczba += p[m] * j;
            }
            s2.insert(liczba);
          }
    int t;
    cin >> t;
    while (t--) solve();
  }
  