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;
  int cases(1), t;
  const int64_t Mod = 1e9 + 7, Mod1 = 998244353;
  template <typename T>
  T Power(T n, T k) {
    T x = 1;
    while (k) {
      if (k & 1) x *= n;
      n *= n;
      k >>= 1;
    }
    return x;
  }
  template <typename T>
  T Power(T n, T k, T M) {
    n %= M;
    T res = 1;
    while (k) {
      if (k & 1) res = res * n % M;
      n = n * n % M;
      k >>= 1;
    }
    return res;
  }
  string one(string s) {
    string ans;
    for (char i = '9'; i >= '0'; i--) {
      string t(s.size(), i);
      if (t >= s) {
        ans = t;
      }
    }
    return ans;
  }
  string two(string s) {
    string ans = one(s);
    for (char a = '0'; a <= '9'; a++) {
      for (char b = a + 1; b <= '9'; b++) {
        bool okay = false;
        for (long long i = 0; i < s.size(); i++) {
          if (s[i] < b) {
            string t = s;
            if (s[i] < a) {
              t[i] = a;
            } else
              t[i] = b;
            for (long long j = i + 1; j < s.size(); j++) {
              t[j] = a;
            }
            if (ans > t) ans = t;
          }
          if (s[i] != a && s[i] != b) {
            okay = true;
            break;
          }
        }
        if (!okay) return s;
      }
    }
    return ans;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(15);
    for (cin >> t; cases <= t; cases++) {
      string n;
      long long k;
      cin >> n >> k;
      if (k == 1)
        cout << one(n);
      else
        cout << two(n);
      cout << '\n';
    }
    return 0;
  }
  