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;
  int t, n, k;
  int a[20], al;
  int b[20], bl;
  int d[20];
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d %d", &n, &k);
      memset(d, 0, sizeof(d));
      memset(a, 0, sizeof(a));
      memset(b, -1, sizeof(b));
      al = 0;
      while (n) {
        a[al] = n % 10;
        n /= 10;
        al++;
      }
      a[al] = 0;
      bl = al;
      int l = bl - 1, c = 0;
      while (l != -1) {
        if (d[a[l]] == 0) {
          c++;
          if (c > k) {
            c = k;
            break;
          }
        }
        d[a[l]]++;
        b[l] = a[l];
        l--;
      }
      if (l != -1) {
        bool tf = false;
        while (1) {
          for (int i = a[l] + 1; i <= 9; i++) {
            if ((d[i] != 0) || (c < k)) {
              if (d[i] == 0) c++;
              b[l] = i;
              d[i]++;
              tf = true;
              break;
            }
          }
          if (tf) break;
          l++;
          d[a[l]]--;
          if (d[a[l]] == 0) c--;
        }
        int m = 0;
        if (c == k) {
          for (int i = 0; i <= 9; i++)
            if (d[i] != 0) {
              m = i;
              break;
            }
        }
        for (int i = l - 1; i >= 0; i--) b[i] = m;
      }
      for (int i = bl - 1; i >= 0; i--) printf("%d", b[i]);
      printf("\n");
    }
    return 0;
  }
  