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 t, n, k;
  int f() {
    int cnt = to_string(n).size();
    if (k == 1) {
      int ans = 0;
      for (int i = 0; i < cnt; i++) ans = ans * 10 + 1;
      for (int i = 0; i < 10; i++)
        if (i * ans >= n) return i * ans;
    }
    int ans = 2e9;
    int s = to_string(n)[0] - '0';
    for (int i = 0; i <= 1 << cnt; i++)
      for (int j = s; j < s + 1; j++)
        for (int k = 0, x, m, cur; k < 10; k++) {
          for (m = i, x = cur = 0; x < cnt; x++, m /= 2)
            cur = cur * 10 + (m % 2 ? j : k);
          if (cur >= n) ans = min(cur, ans);
        }
    return ans;
  }
  signed main() {
    for (cin >> t; t--;) {
      cin >> n >> k;
      cout << f() << "\n";
    }
  }
  