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;
  string n;
  int k;
  long long int powers[11];
  long long int solve(int curDigits, int mask, bool bigger) {
    if (curDigits < 0) return 0;
    int remK = k - __builtin_popcount(mask);
    long long int ans = 1e15;
    for (int i = (bigger ? 0 : n[curDigits] - '0'); i <= 9; i++) {
      int newState = bigger;
      if (i > n[curDigits] - '0') newState = true;
      if (mask & (1 << i)) {
        long long int tmp = solve(curDigits - 1, mask, newState);
        if (tmp < 1e15) {
          ans = i * powers[curDigits] + tmp;
          break;
        }
      } else {
        if (remK > 0) {
          long long int tmp = solve(curDigits - 1, mask | (1 << i), newState);
          if (tmp < 1e15) {
            ans = i * powers[curDigits] + tmp;
            break;
          }
        }
      }
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    powers[0] = 1;
    for (int i = 1; i < 10; i++) powers[i] = powers[i - 1] * 10;
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      reverse(n.begin(), n.end());
      cout << solve(n.size() - 1, 0, false) << "\n";
    }
    return 0;
  }
  