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;
  string numbertoString(long long int n) {
    string s = "";
    while (n > 0) {
      long long int d = n % 10;
      char ch = '0' + d;
      s += ch;
      n /= 10;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  long long int binexp(long long int x, long long int n) {
    if (n == 0)
      return 1ll;
    else if (n & 1ll) {
      return (x * binexp(x, n - 1));
    } else {
      long long int y = binexp(x, n / 2);
      return (y * y);
    }
  }
  long long int stringtonumber(string s) {
    long long int num = 0;
    long long int len = s.length();
    for (long long int i = 0; i < s.length(); i++) {
      long long int temp = binexp(10, len - i - 1) * (s[i] - '0');
      num += temp;
    }
    return num;
  }
  void Helper(long long int n, long long int k) {
    string s = numbertoString(n);
    vector<long long int> vec;
    if (k == 2) {
      for (long long int a = 0; a < 10; a++) {
        for (long long int b = 0; b < 10; b++) {
          bool ok = false;
          string t = "";
          long long int mn = min(a, b), mx = max(a, b);
          for (long long int pos = 0; pos < s.length(); pos++) {
            if (ok) {
              char ch = '0' + mn;
              t += ch;
            } else {
              if (mn >= (s[pos] - '0')) {
                long long int digs_left = s.length() - pos - 1;
                string left = "";
                char ch = '0' + mn;
                left += ch;
                for (long long int times = 0; times < digs_left; times++) {
                  ch = '0' + mx;
                  left += ch;
                }
                long long int mx_pos = 0;
                if (!left.empty()) mx_pos = stringtonumber(left);
                string gus = "";
                if (pos <= (n - 1)) gus = s.substr(pos, digs_left + 1);
                long long int shawn = 0;
                if (!gus.empty()) shawn = stringtonumber(gus);
                if (shawn > mx_pos) {
                  ch = '0' + mx;
                  t += ch;
                  long long int cur = s[pos] - '0';
                  if (mx > cur) ok = true;
                } else {
                  char ch = '0' + mn;
                  t += ch;
                  long long int cur = s[pos] - '0';
                  if (mn > cur) ok = true;
                }
              } else {
                char ch = '0' + mx;
                t += ch;
                if (mx > (s[pos] - '0')) ok = true;
              }
            }
          }
          long long int val = stringtonumber(t);
          if (val >= n) vec.push_back(val);
        }
      }
    } else if (k == 1) {
      for (long long int a = 0; a < 10; a++) {
        string t = "";
        for (long long int j = 0; j < s.length(); j++) {
          char ch = '0' + a;
          t += ch;
        }
        long long int num = stringtonumber(t);
        if (num >= n) vec.push_back(num);
      }
    }
    long long int ans = *min_element(vec.begin(), vec.end());
    cout << ans << "\n";
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n, k;
      cin >> n >> k;
      Helper(n, k);
    }
  }
  