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 long long INF = 1e18 + 7;
  const int inf = 0x3f3f3f3f;
  const int N = 5e5 + 500;
  const int MAXN = 5e5 + 5;
  const int mod = 1e9 + 7;
  const int md = 1e9 + 7;
  const int MOD = 100003;
  const int Mod = 998244353;
  const int M = 30;
  const long long moD = 100000;
  const int D = 233;
  long long powmod(long long a, long long b) {
    long long res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
    }
    return res;
  }
  long long gcd(long long a, long long b) { return (!b ? a : gcd(b, a % b)); }
  int T;
  long long k;
  string n;
  string solve1(string s) {
    string res(s.size(), '9');
    for (int i = 8; i >= 0; i--) {
      string t(s.size(), char(i + '0'));
      if (t >= s) res = t;
    }
    return res;
  }
  string solve2(string s) {
    string res = solve1(s);
    for (char a = '0'; a <= '8'; a++) {
      for (char b = a + 1; b <= '9'; b++) {
        int flag = 1;
        for (int i = 0; i < s.size(); i++) {
          if (s[i] < b) {
            string t = s;
            if (t[i] < a)
              t[i] = a;
            else
              t[i] = b;
            for (int k = i + 1; k < (int)s.size(); k++) {
              t[k] = a;
            }
            if (t >= s) res = min(res, t);
          }
          if (s[i] != a && s[i] != b) {
            flag = 0;
            break;
          }
        }
        if (flag) return n;
      }
    }
    return res;
  }
  int main() {
    cin >> T;
    while (T--) {
      cin >> n >> k;
      if (k == 1)
        cout << solve1(n) << endl;
      else
        cout << solve2(n) << endl;
    }
    return 0;
  }
  