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>
  #pragma GCC optimize("Ofast,fast-math,unroll-loops")
  #pragma GCC target("avx,avx2,sse,sse2,sse3,ssse3,sse4,abm,mmx,popcnt")
  using namespace std;
  const long long INF = 1e18 + 13;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << H;
    debug_out(T...);
  }
  template <typename T>
  std::istream& operator>>(std::istream& in, std::vector<T> vec) {
    for (T& a : vec) in >> a;
    return in;
  }
  const string SEPARATOR = " ";
  template <typename T>
  std::ostream& operator<<(std::ostream& out, std::vector<T> vec) {
    for (T& a : vec) out << a << SEPARATOR;
    return out;
  }
  template <typename T, typename U>
  std::istream& operator>>(std::istream& in, std::pair<T, U> pr) {
    in >> pr.first >> pr.second;
    return in;
  }
  template <typename T, typename U>
  std::ostream& operator<<(std::ostream& out, std::pair<T, U> pr) {
    out << pr.first << SEPARATOR << pr.second;
    return out;
  }
  long long k1sz = 0;
  long long nxt = 0;
  long long k1[100], k2[1000000];
  void solve() {
    long long n, k;
    cin >> n >> k;
    if (k == 1) {
      cout << *lower_bound(k1, k1 + k1sz, n);
    } else
      cout << *lower_bound(k2, k2 + nxt, n);
    cout << endl;
  }
  void rec(string& n, long long k, long long l) {
    k2[nxt++] = stoll(n);
    if (((long long)(n).size()) == 10) return;
    n.push_back((char)(k + '0'));
    rec(n, k, l);
    n.pop_back();
    n.push_back((char)(l + '0'));
    rec(n, k, l);
    n.pop_back();
  }
  void precalc() {
    for (long long i = 1; i < 10; ++i) {
      string s = "";
      for (long long j = 1; j <= 10; ++j) {
        s.push_back((char)(i + '0'));
        k1[k1sz++] = stoll(s);
      }
    }
    sort(k1, k1 + k1sz);
    string s;
    for (long long i = 1; i < 10; ++i) {
      s.push_back((char)(i + '0'));
      for (long long j = 0; j < 10; ++j) {
        rec(s, i, j);
      }
      s.pop_back();
    }
    sort(k2, k2 + nxt);
  }
  signed main() {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    long long n = 1;
    cin >> n;
    precalc();
    while (n--) {
      solve();
    }
  }
  