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;
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  const int MXN = 1e5 + 5, INF = 1e9 + 5;
  void solve() {
    string S;
    int K;
    cin >> S >> K;
    int N = (int)(S).size();
    vector<int> A(N), cnt(10, 0);
    int distinct = 0;
    for (int i = 0; i < N; i++) {
      A[i] = S[i] - '0';
      cnt[A[i]]++;
      if (cnt[A[i]] == 1) distinct++;
    }
    if (distinct <= K) {
      cout << S << "\n";
      return;
    }
    vector<int> candidates;
    for (int i = N - 1; i >= 0; i--) {
      int orig = A[i];
      cnt[orig]--;
      if (cnt[orig] == 0) distinct--;
      for (int inc = A[i] + 1; inc <= 9; inc++) {
        cnt[inc]++;
        distinct += (cnt[inc] == 1);
        for (int j = i + 1; j < N; j++) {
          cnt[A[j]]--;
          distinct -= (cnt[A[j]] == 0);
        }
        int mn = 9;
        for (int j = 0; j < 10; j++) {
          if (cnt[j]) {
            mn = j;
            break;
          }
        }
        if (distinct <= K) {
          for (int j = i + 1; j < N; j++) A[j] = (distinct < K ? 0 : mn);
          A[i] = inc;
          for (const auto &x : A) cout << x;
          cout << "\n";
          return;
        }
        for (int j = i + 1; j < N; j++) {
          cnt[A[j]]++;
          distinct += (cnt[A[j]] == 1);
        }
        cnt[inc]--;
        distinct -= (cnt[inc] == 0);
      }
      cnt[orig]++;
      if (cnt[orig] == 1) distinct++;
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int TC = 1;
    cin >> TC;
    while (TC--) solve();
  }
  