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 long long INF = 5e9;
  int t, n, k, lg10[10];
  int digit(int num, int pos) { return (num / lg10[pos]) % 10; }
  int logarithm(int num) {
    int ret = 0;
    while (num > 0) {
      num /= 10;
      ret++;
    }
    return ret - 1;
  }
  long long memo[11][1 << 10][2];
  long long rek(int pos, int mask, int f) {
    if (__builtin_popcount(mask) > k) return INF;
    if (pos == -1) return 0LL;
    long long &ret = memo[pos][mask][f];
    if (ret != -1) return ret;
    int dig = digit(n, pos);
    if (f) {
      for (int i = 0; i < 10; i++) {
        if (rek(pos - 1, mask | (1 << i), 1) < INF)
          return ret =
                     rek(pos - 1, mask | (1 << i), 1) + (long long)i * lg10[pos];
      }
      return ret = INF;
    }
    if (rek(pos - 1, mask | (1 << dig), 0) < INF)
      return ret =
                 rek(pos - 1, mask | (1 << dig), 0) + (long long)dig * lg10[pos];
    for (int i = dig + 1; i < 10; i++) {
      if (rek(pos - 1, mask | (1 << i), 1) < INF)
        return ret = rek(pos - 1, mask | (1 << i), 1) + (long long)i * lg10[pos];
    }
    return ret = INF;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    lg10[0] = 1;
    for (int i = 1; i < 10; i++) lg10[i] = lg10[i - 1] * 10;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      memset(memo, -1LL, sizeof memo);
      cout << rek(logarithm(n), 0, 0) << '\n';
    }
    return 0;
  }
  