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;
  int a[150];
  int vis[10];
  long long ans, len;
  string s;
  void dfs(int p, int k, long long n, int f) {
    if (k == -1) return;
    if (p >= len) {
      ans = min(ans, n);
      return;
    }
    if (f) {
      int mn;
      if (k)
        mn = 0;
      else {
        for (int i = 0; i < 10; i++) {
          if (vis[i]) {
            mn = i;
            break;
          }
        }
      }
      for (int i = p; i < len; i++) n = n * 10 + mn;
      ans = min(ans, n);
      return;
    }
    ++vis[s[p] - '0'];
    dfs(p + 1, k - (vis[s[p] - '0'] == 1), n * 10 + s[p] - '0', 0);
    --vis[s[p] - '0'];
    if (s[p] != '9') {
      ++vis[s[p] - '0' + 1];
      dfs(p + 1, k - (vis[s[p] - '0' + 1] == 1), n * 10 + s[p] - '0' + 1, 1);
      --vis[s[p] - '0' + 1];
    }
    int fd = 0;
    for (int i = s[p] - '0' + 2; i < 10; i++) {
      if (vis[i]) {
        fd = i;
        break;
      }
    }
    if (fd) {
      ++vis[fd];
      dfs(p + 1, k, n * 10 + fd, 1);
      --vis[fd];
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      s = to_string(n);
      len = s.size();
      ans = 1111111111;
      dfs(0, k, 0, 0);
      cout << ans << '\n';
    }
    return 0;
  }
  