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 oo = 1e18;
  const int MX = INT_MAX;
  const int MN = INT_MIN;
  const int N = 3e5 + 5;
  const int mod = 1e9 + 7;
  vector<long long> v, v1;
  void build(long long val, long long i, long long x, long long y) {
    if (1) {
      v.push_back(val);
    }
    if (i > 11) return;
    long long ret = val + (x * pow(10, i));
    long long ret1 = val + (y * (pow(10, i)));
    build(ret1, i + 1, x, y);
    build(ret, i + 1, x, y);
  }
  int main() {
    for (int i = 0; i < 10; i++) {
      for (int j = i + 1; j < 10; j++) build(0, 0, i, j);
    }
    for (int i = 1; i < 10; i++) {
      long long val = i;
      while (val <= 1e11) {
        v1.push_back(val);
        val *= 10;
        val += i;
      }
    }
    sort(v.begin(), v.end());
    sort(v1.begin(), v1.end());
    bool f = 0;
    int test_cases = 1;
    cin >> test_cases;
    while (test_cases--) {
      long long x, k;
      cin >> x >> k;
      if (k == 1) {
        int s = 0, e = v1.size() - 1, mid;
        long long val;
        while (s <= e) {
          mid = (s + e) / 2;
          if (v1[mid] >= x) {
            val = v1[mid];
            e = mid - 1;
          } else
            s = mid + 1;
        }
        cout << val << "\n";
      } else {
        int s = 0, e = v.size() - 1, mid;
        long long val;
        while (s <= e) {
          mid = (s + e) / 2;
          if (v[mid] >= x) {
            val = v[mid];
            e = mid - 1;
          } else
            s = mid + 1;
        }
        cout << val << "\n";
      }
    }
    return 0;
  }
  