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;
  long long n, k;
  string s;
  long long fun(long long idx, long long nn, set<long long> &st, long long f) {
    if (st.size() > k) return 1e15;
    if (nn >= n) return nn;
    long long ans = 1e15;
    if (f) {
      if (st.size() < k) {
        ans = fun(idx + 1, nn * 10, st, 1);
      } else {
        ans = fun(idx + 1, nn * 10 + *st.begin(), st, 1);
      }
    } else {
      long long l, r;
      if (idx >= s.size()) {
        l = 0, r = 9;
      } else {
        l = s[idx] - '0';
        r = 9;
      }
      long long i = l;
      for (; l <= r; ++l) {
        long long cur;
        if (st.count(l)) {
          cur = fun(idx + 1, nn * 10 + l, st, f || l > i);
        } else {
          st.insert(l);
          cur = fun(idx + 1, nn * 10 + l, st, f || l > i);
          st.erase(l);
        }
        ans = min(ans, cur);
      }
    }
    return ans;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      s = to_string(n);
      set<long long> tt;
      cout << fun(0, 0, tt, 0) << endl;
    }
    return 0;
  }
  