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 N = 1000000007;
  vector<long long> v, y;
  void solve() {
    for (long long i = 1; i <= 9; i++) {
      for (long long l = 1; l <= 10; l++) {
        long long t = 0;
        for (long long k = 0; k < l; k++) {
          t = t * 10 + i;
        }
        y.push_back(t);
      }
    }
    for (long long i = 0; i <= 9; i++) {
      for (long long j = i + 1; j <= 9; j++) {
        for (long long l = 1; l <= 10; l++) {
          for (long long mask = 0; mask < (1 << l); mask++) {
            long long num = 0;
            for (long long k = 0; k < l; k++) {
              bool y = ((1 << k) & mask);
              if (y) {
                num = num * 10 + i;
              } else {
                num = num * 10 + j;
              }
            }
            v.push_back(num);
          }
        }
      }
    }
    sort(y.begin(), y.end());
    sort(v.begin(), v.end());
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t;
    cin >> t;
    solve();
    while (t--) {
      long long n, k;
      cin >> n >> k;
      if (k == 1) {
        long long l = 0, r = y.size() - 1, mid = 0;
        while (l < r) {
          mid = l + (r - l) / 2;
          if (y[mid] < n) {
            l = mid + 1;
          } else {
            r = mid;
          }
        }
        cout << y[l] << "\n";
      } else {
        long long l = 0, r = v.size() - 1, mid = 0;
        while (l < r) {
          mid = l + (r - l) / 2;
          if (v[mid] < n) {
            l = mid + 1;
          } else {
            r = mid;
          }
        }
        cout << v[l] << "\n";
      }
    }
    return 0;
  }
  