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;
  int bi[20];
  int ok(long long p) {
    int f[10] = {0}, op = 0;
    for (long long o = p; o; o /= 10) f[(int)(o % 10)] = 1;
    for (int i = 0; i < 10; i++) op += f[i];
    return op;
  }
  int f[20];
  int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    int tc;
    cin >> tc;
    while (tc--) {
      cin >> n >> k;
      int r = 0;
      if (ok(n) <= k) {
        cout << n << endl;
        continue;
      }
      for (int i = n; i; i /= 10) bi[r++] = i % 10;
      int sp = 0;
      for (long long i = n; i; i /= 10, sp++) {
        long long rr = i % 10, p = i / 10;
        int ss = -1;
        for (int j = rr + 1; j < 10; j++)
          if (ok(p * 10 + j) <= k) {
            ss = j;
            break;
          }
        if (ss > -1) {
          long long io = p * 10 + ss;
          memset(f, 0, sizeof(f));
          int sq;
          for (long long j = io; j; j /= 10) f[j % 10] = 1;
          for (int j = 0; j < 10; j++)
            if (f[j]) {
              sq = j;
              break;
            }
          if (ok(io) < k) sq = 0;
          for (int j = 0; j < sp; j++) io = io * 10 + sq;
          cout << io << endl;
          break;
        }
      }
    }
  }
  