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;
  using ll = long long;
  string n;
  void solve1() {
    ll len = n.size();
    for (char c = '1'; c <= '9'; ++c) {
      string ans = "";
      for (ll i = 1; i <= len; ++i) ans += c;
      if (ans >= n) {
        cout << ans << '\n';
        return;
      }
    }
  }
  void solve2() {
    ll len = n.size();
    string res;
    for (char c = '1'; c <= '9'; ++c) {
      string ans = "";
      for (ll i = 1; i <= len; ++i) ans += c;
      if (ans >= n) {
        res = ans;
        break;
      }
    }
    for (char a = '0'; a <= '9'; ++a) {
      for (char b = a + 1; b <= '9'; ++b) {
        bool ok = 1;
        for (ll i = 0; i < len; ++i) {
          if (n[i] < b) {
            string ans = n;
            if (n[i] < a)
              ans[i] = a;
            else
              ans[i] = b;
            for (ll j = i + 1; j < len; j++) ans[j] = a;
            if (ans < res) res = ans;
          }
          if (n[i] != a && n[i] != b) {
            ok = 0;
            break;
          }
        }
        if (ok) {
          cout << n << '\n';
          return;
        }
      }
    }
    cout << res << '\n';
    return;
  }
  void solve() {
    ll k;
    cin >> n;
    cin >> k;
    if (k == 1)
      solve1();
    else
      solve2();
  }
  signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    ll t;
    cin >> t;
    while (t--) solve();
  }
  