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>
  #pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
  using namespace std;
  vector<int> digits;
  int k, cur_k;
  bool used[10];
  int64_t dfs(int pos, int64_t prefix, bool same) {
    if (pos == digits.size()) return prefix;
    int tried = 0;
    int64_t ans = INT_MAX;
    for (int i = 0; i < 10; ++i) {
      if (tried == 2) break;
      if (same && i < digits[pos]) continue;
      if (k == cur_k && !used[i]) continue;
      bool is_new = used[i];
      if (!is_new) {
        used[i] = true;
        ++cur_k;
      }
      ++tried;
      ans = min(ans, dfs(pos + 1, prefix * 10 + i, same && i == digits[pos]));
      if (!is_new) {
        used[i] = false;
        --cur_k;
      }
    }
    return ans;
  }
  void solve() {
    int64_t n;
    cin >> n >> k;
    digits.clear();
    cur_k = 0;
    memset(used, 0, sizeof(used));
    int64_t temp = n;
    while (temp) {
      digits.push_back(temp % 10);
      temp /= 10;
    }
    reverse(digits.begin(), digits.end());
    int64_t ans = dfs(0, 0, true);
    cout << ans << "\n";
    return;
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  