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;
  const double pai = acos(-1.0);
  const int maxn = 1e6 + 10;
  const int mod = 1e9 + 7;
  const double eps = 1e-9;
  const int N = 5e3 + 10;
  inline int read() {
    int k = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (isdigit(c)) k = (k << 1) + (k << 3) + c - 48, c = getchar();
    return k * f;
  }
  int t, n, k, cnt[1030];
  int a[20];
  long long dp[20][1030][12];
  long long dfs(int pos, int pre, int state, bool lead, bool limit) {
    if (pos == -1 && cnt[state] <= k) return 1;
    if (!limit && !lead && dp[pos][state][k] != -1) return dp[pos][state][k];
    int up = limit ? a[pos] : 9;
    long long res = 0;
    for (int i = 0; i <= up; i++) {
      if (i == 0 && lead)
        res += dfs(pos - 1, i, state, lead && i == 0, limit && i == a[pos]);
      else {
        if (cnt[state | (1 << i)] > k) continue;
        res += dfs(pos - 1, i, (state | (1 << i)), lead && i == 0,
                   limit && i == a[pos]);
      }
    }
    if (!limit && !lead) dp[pos][state][k] = res;
    return res;
  }
  long long solve(int x) {
    int pos = 0;
    while (x) {
      a[pos++] = x % 10;
      x /= 10;
    }
    return dfs(pos - 1, -1, 0, 1, 1);
  }
  int main() {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i < 1 << 10; i++) cnt[i] = __builtin_popcount(i);
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &k);
      long long num = solve(n - 1);
      long long l = n, r = 2e9, ans = 0;
      while (l <= r) {
        int mid = l + r >> 1;
        if (solve(mid) - num >= 1) {
          r = mid - 1;
          ans = mid;
        } else
          l = mid + 1;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  