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(1)
  #pragma GCC optimize(2)
  #pragma GCC optimize(3)
  using namespace std;
  const int mo = 998244353;
  const int maxn = 2e5 + 10;
  int t = 1, n, m, k;
  long long num1[maxn], num2[maxn];
  int cnt1 = 0, cnt2 = 0;
  void dfs(int step, long long num, int x, int y) {
    num2[cnt2++] = num;
    if (step == 10) return;
    dfs(step + 1, num * 10 + x, x, y);
    dfs(step + 1, num * 10 + y, x, y);
  }
  int main() {
    scanf("%d", &t);
    for (int i = 1; i <= 9; i++) {
      long long num = 0;
      for (int j = 1; j <= 10; j++) {
        num *= 10;
        num += i;
        num1[cnt1++] = num;
      }
    }
    sort(num1, num1 + cnt1);
    cnt2 = cnt1;
    memcpy(num2, num1, sizeof(num1));
    for (int i = 0; i <= 8; i++)
      for (int j = i + 1; j <= 9; j++) {
        dfs(0, 0, i, j);
      }
    sort(num2, num2 + cnt2);
    while (t--) {
      scanf("%d%d", &n, &k);
      if (k == 1) {
        printf("%d\n", num1[lower_bound(num1, num1 + cnt1, n) - num1]);
      } else {
        printf("%d\n", num2[lower_bound(num2, num2 + cnt2, n) - num2]);
      }
    }
  }
  