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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  long long st[10][1 << 10];
  void solve() {
    int n, kkk;
    scanf("%d%d", &n, &kkk);
    int x = n, len = 0;
    while (x) x /= 10, len++;
    if (kkk == 1) {
      for (int num = 1; num <= 9; num++) {
        long long sum = 0, cur = 1;
        for (int i = 1; i <= len; i++) sum += num * cur, cur *= 10;
        if (sum >= n) {
          printf("%lld\n", sum);
          return;
        }
      }
    } else {
      long long ans = 1e18;
      for (int num1 = 0; num1 <= 9; num1++)
        for (int num2 = num1; num2 <= 9; num2++)
          for (int s = 0; s < (1 << len); s++) {
            long long sum = st[num1][s] + st[num2][(~s) & ((1 << len) - 1)];
            if (sum >= n) ans = min(ans, sum);
          }
      printf("%lld\n", ans);
    }
  }
  int main() {
    for (int num = 0; num <= 9; num++)
      for (int s = 0; s < (1 << 10); s++) {
        long long sum = 0, cur = 1;
        for (int i = 1; i <= 10; i++)
          sum += ((s & (1 << (i - 1))) ? num : 0) * cur, cur *= 10;
        st[num][s] = sum;
      }
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  