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 MAXN = 40000;
  const int MOD = 1e6;
  string str;
  string solve() {
    int k;
    cin >> str >> k;
    if (k == 1) {
      if (string(str.size(), str[0]) >= str)
        return string(str.size(), str[0]);
      else
        return string(str.size(), str[0] + 1);
    } else {
      char ch1 = str[0];
      bool fg;
      string ans, res = string(str.size(), '9');
      for (char ch2 = '0'; ch2 <= '9'; ++ch2) {
        fg = 0;
        ans.clear();
        ans += ch1;
        char mi = min(ch2, ch1), ma = max(ch2, ch1);
        int i;
        for (i = 1; i < str.size(); ++i) {
          if (mi > str[i]) {
            ans += string(str.size() - i, mi);
            fg = 1;
            break;
          } else if (mi == str[i]) {
            ans += mi;
          } else {
            if (ma > str[i]) {
              ans += ma;
              ans += string(str.size() - i - 1, mi);
              fg = 1;
              break;
            } else if (ma == str[i]) {
              ans += ma;
            } else {
              if (ma == mi)
                break;
              else if (i == 1) {
                break;
              } else {
                ans += string(str.size() - i, mi);
                for (int j = i - 1; j >= 1; --j) {
                  if (ans[j] == mi) {
                    ans[j] = ma;
                    fg = 1;
                    break;
                  } else
                    ans[j] = mi;
                }
                break;
              }
            }
          }
        }
        if ((fg || i == str.size()) && ans < res) res = ans;
      }
      return res;
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      cout << solve() << '\n';
    }
    return 0;
  }
  