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 = 1e6 + 7;
  const double eps = 1e-6;
  const int mod = 1e9 + 7;
  int a[N];
  int digits(int n) {
    bool f[10] = {0};
    while (n) {
      f[n % 10] = 1;
      n /= 10;
    }
    int cnt = 0;
    for (int i = 0; i < 10; i++) cnt += f[i];
    return cnt;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, k, t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      while (digits(n) > k) {
        int r = 1;
        int c = n;
        while (digits(c) > k) {
          c /= 10;
          r *= 10;
        }
        r /= 10;
        n = ((n / r) + 1) * r;
      }
      cout << n << endl;
    }
    return 0;
  }
  