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 k, t, p[100010], d;
  long long minn, maxn, n;
  int dig(int x) {
    int ans = 0;
    while (x) {
      ans++;
      p[ans] = x % 10;
      maxn = max(maxn, p[ans]);
      x = x / 10;
    }
    return ans;
  }
  void dfs(int id, int x, long long sum) {
    if (id == 1) {
      if (sum >= n) minn = min(minn, sum);
      return;
    }
    dfs(id - 1, x, sum * 10 + x);
    dfs(id - 1, x, sum * 10 + p[d]);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      minn = 1e11;
      maxn = 0;
      d = dig(n);
      if (k == 1) {
        long long s;
        for (int i = 1; i <= 9; i++) {
          s = 0;
          for (int j = 1; j <= d; j++) {
            s = s * 10 + i;
          }
          if (s >= n) minn = min(minn, s);
        }
        cout << minn << "\n";
        continue;
      }
      for (int i = 0; i <= 9; i++) dfs(d, i, p[d]);
      cout << minn << "\n";
    }
  }
  