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;
  set<int> used;
  int k;
  string digs;
  int pre;
  int go(int rem, bool match) {
    if (rem == 0) return pre;
    int p = digs.length() - rem;
    int tried = 0;
    for (int i = 0; i < 10; i++) {
      if (match and i < (int)digs[p] - '0') continue;
      if (used.size() == k and used.count(i) == 0) continue;
      if (tried >= 2) continue;
      tried++;
      pre = 10 * pre, pre += i;
      bool curin = used.count(i);
      if (!curin) used.insert(i);
      int res = go(rem - 1, (match and i == digs[p] - '0'));
      if (!curin) used.erase(i);
      pre /= 10;
      if (res != -1) {
        return res;
      }
    }
    return -1;
  }
  void solve() {
    int n;
    cin >> n >> k;
    used.clear();
    pre = 0;
    digs = to_string(n);
    int res = go(digs.length(), true);
    cout << res << endl;
  }
  int main() {
    int t;
    cin >> t;
    while (t) {
      solve();
      t--;
    }
  }
  