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 MAX_N = 5e5 + 5;
  const int INF = 0x3f3f3f3f;
  const int mod = 1e9 + 7;
  const double eps = 1e-6;
  int n, k;
  long long sum(long long first) {
    long long ans = 0;
    while (first) {
      if (first & 1) ans++;
      first >>= 1;
    }
    return ans;
  }
  long long calc(long long first) {
    long long p = 0;
    while (first) {
      p |= (1 << (first % 10));
      first /= 10;
    }
    return sum(p);
  }
  void solve() {
    cin >> n >> k;
    while (calc(n) > k) {
      long long p = n;
      long long r = 1;
      while (calc(p) > k) {
        p /= 10;
        r *= 10;
      }
      r /= 10;
      n = ((n / r) + 1) * r;
    }
    cout << n << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    int test = 1;
    cin >> test;
    while (test--) {
      solve();
    }
    return 0;
  }
  