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 int N = 1000000;
  long long int M = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
    long long int t = 1, i, j, num;
    cin >> t;
    while (t--) {
      long long int n, k, m, cnt = 1;
      cin >> n >> k;
      string ss = to_string(n);
      if (k == 1 && n < 10) {
        cout << n << "\n";
        continue;
      }
      for (i = 1; i < ss.size(); i++) {
        if (ss[i] == ss[0]) {
          cnt++;
        }
      }
      if (cnt == ss.size() && k == 2) {
        cout << n << "\n";
        continue;
      }
      if (k == 1) {
        string s = to_string(n);
        num = s.size();
        string kk(num, '1');
        long long int n1 = stoll(kk);
        long long int k2 = ceil((long double)n / n1);
        long long int ans = k2 * n1;
        cout << ans << "\n";
        continue;
      } else if (k == 2) {
        long long int ans = 0, diff = 1e13;
        string s = to_string(n);
        for (i = 0; i < 10; i++) {
          for (j = i + 1; j < 10; j++) {
            char c1 = '0' + i;
            char c2 = '0' + j;
            string anu = "";
            m = 0;
            long long int flg = 0, fllg = 0, loc = -1;
            for (m = 0; m < s.size(); m++) {
              if (flg == 1 || s[m] == c1) {
                anu += c1;
                fllg = 1;
                loc = m;
              } else if (s[m] < c1) {
                anu += c1;
                flg = 1;
              } else if (s[m] == c2) {
                anu += c2;
              } else if (s[m] > c2) {
                if (fllg == 0) {
                  break;
                } else {
                  long long int n6 = loc;
                  anu[n6] = c2;
                  for (long long int kk1 = n6 + 1; kk1 < m; kk1++) {
                    anu[kk1] = c1;
                  }
                  for (long long int kk1 = m; kk1 < s.size(); kk1++) {
                    anu += c1;
                  }
                  break;
                }
              } else if (s[m] > c1 && s[m] < c2) {
                anu += c2;
                flg = 1;
              }
            }
            if (anu.size() == 0) {
              continue;
            }
            long long int n3 = stoll(anu);
            if (n3 < n) {
              continue;
            }
            long long int n4 = n3 - n;
            if (n4 < diff) {
              diff = n4;
              ans = n3;
            }
          }
        }
        cout << ans << "\n";
        continue;
      }
    }
  }
  