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;
  set<int> se;
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, k;
      se.clear();
      scanf("%d%d", &n, &k);
      int mark = n;
      int b[20];
      int tcnt = 0;
      while (n) {
        b[++tcnt] = n % 10;
        n /= 10;
      }
      int vis[10] = {0};
      int nowcnt = 0;
      int i;
      int markstart;
      for (i = tcnt; i >= 1; i--) {
        if (vis[b[i]] == 0) {
          nowcnt++;
          vis[b[i]] = 1;
          if (nowcnt <= k) se.insert(b[i]);
          if (nowcnt == k) markstart = i;
        }
        if (nowcnt > k) {
          if (b[i] > (*(--se.end()))) {
            for (int j = i + 1; j < markstart; j++) {
              if (b[j] < (*(--se.end()))) {
                b[j] = (*se.upper_bound(b[j]));
                i = j - 1;
                goto endd2;
              }
            }
            goto endd;
          } else {
            b[i] = (*se.lower_bound(b[i]));
            i--;
            break;
          }
        }
      }
    endd2:;
      for (; i >= 1; i--) b[i] = (*se.begin());
      if (nowcnt <= k)
        printf("%d\n", mark);
      else {
        for (int i = tcnt; i >= 1; i--) printf("%d", b[i]);
        cout << endl;
      }
      continue;
    endd:;
      se.clear();
      memset(vis, 0, sizeof(vis));
      nowcnt = 0;
      for (i = tcnt; i >= 1; i--) {
        if (vis[b[i]] == 0) {
          nowcnt++;
          vis[b[i]] = 1;
          if (nowcnt < k) se.insert(b[i]);
        }
        if (nowcnt == k) {
          se.insert(b[i] + 1);
          if (se.size() < k) se.insert(0);
          b[i]++;
          i--;
          break;
        }
      }
      for (; i >= 1; i--) b[i] = (*se.begin());
      for (int i = tcnt; i >= 1; i--) printf("%d", b[i]);
      cout << endl;
    }
    return 0;
  }
  