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;
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &p) {
    os << "[";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << "(" << p.first << "," << p.second << ")";
  }
  template <class T>
  bool umin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool umax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  void stugs() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long int n, k;
  string str;
  long long int fool[12][1030][2];
  long long int dp(long long int i, long long int mask, long long int d,
                   string &ans) {
    if (__builtin_popcount(mask) > k) return 0;
    if (i >= n) return 1;
    if (fool[i][mask][d] != -1) return fool[i][mask][d];
    char lb = '0';
    if (d == 1) lb = str[i];
    for (char j = lb; j <= '9'; j++) {
      long long int x = 0;
      if (d == 1 and j == lb) x = 1;
      if (dp(i + 1, mask | (1 << (j - '0')), x, ans)) {
        ans = j + ans;
        return fool[i][mask][d] = 1;
      }
    }
    return fool[i][mask][d] = 0;
  }
  void solve() {
    cin >> str >> k;
    n = str.length();
    string ans = "";
    memset(fool, -1, sizeof fool);
    dp(0, 0, 1, ans);
    cout << ans << endl;
  }
  int32_t main() {
    stugs();
    long long int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  