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 N = 1e5 + 5;
  int n, k;
  int64_t to_i64(string s) {
    int64_t ret = 0;
    for (char ch : s) ret = ret * 10 + (ch - '0');
    return ret;
  }
  int64_t get(string l, string r) {
    set<char> s;
    string ret = l;
    for (char ch : l) s.insert(ch);
    if (s.size() > k) return INT_MAX;
    if (r.empty()) return to_i64(ret);
    if (r[0] == '9') return INT_MAX;
    auto it = s.upper_bound(r[0]);
    if (it == s.end() && s.size() == k) return INT_MAX;
    if (s.size() < k) s.insert(r[0] + 1);
    ret += *s.upper_bound(r[0]);
    set<char> t = s;
    t.insert('0');
    if (t.size() <= k) s = t;
    while (ret.size() < l.size() + r.size()) ret += *s.begin();
    ;
    return to_i64(ret);
  }
  void solve(int Case) {
    scanf("%d %d", &n, &k);
    string s = to_string(n);
    ;
    int64_t ans = INT_MAX;
    ans = min(ans, get("", s));
    ans = min(ans, get(s, ""));
    for (int i = 1; i <= s.size() - 1; ++i)
      ans = min(ans, get(s.substr(0, i), s.substr(i, s.size() - i)));
    if (k >= 2) {
      int64_t tmp = 1;
      while (tmp < n) tmp *= 10;
      ans = min(ans, tmp);
    }
    printf("%lld\n", ans);
  }
  int main() {
    int T = 1;
    scanf("%d", &T);
    for (int t = 1; t <= T; ++t) solve(t);
    return 0;
  }
  