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;
  vector<long long> g[11], f;
  void dfs(long long num, int a, int b, int l, int mdl) {
    if (l == 0) {
      g[mdl].push_back(num);
      return;
    }
    dfs(num * 10 + a, a, b, l - 1, mdl);
    dfs(num * 10 + b, a, b, l - 1, mdl);
  }
  void init() {
    for (int i = 0; i <= 9; i++) {
      long long num = 0;
      for (int j = 1; j <= 10; j++) {
        num = num * 10 + i;
        g[j].push_back(num), f.push_back(num);
      }
    }
    for (int l = 1; l <= 10; l++) {
      for (int i = 0; i <= 9; i++)
        for (int j = i + 1; j <= 9; j++) dfs(0, i, j, l, l);
    }
    for (int l = 1; l <= 10; l++) sort(g[l].begin(), g[l].end());
  }
  int main() {
    init();
    int T;
    cin >> T;
    while (T--) {
      char s[12];
      long long o = 0;
      int k;
      scanf("%s", s + 1);
      int l = strlen(s + 1);
      for (int j = 1; j <= l; j++) o = o * 10 + (s[j] - '0');
      scanf("%d", &k);
      if (k == 1) {
        long long ans = 100000000000ll;
        for (int j = 0; j < f.size(); j++) {
          if (f[j] >= o) ans = min(ans, f[j]);
        }
        printf("%lld\n", ans);
      } else {
        for (int j = 0; j < g[l].size(); j++)
          if (g[l][j] >= o) {
            printf("%lld\n", g[l][j]);
            break;
          }
      }
    }
    return 0;
  }
  