description: It is a complicated version of problem F1. 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 ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long mod = 1e9 + 7;
  long long cl(long double a) {
    if (a > (long long)a) {
      return (long long)a + 1;
    } else {
      return (long long)a;
    }
  }
  long long flr(long double a) {
    if (a < 0.0) {
      return (long long)a - 1;
    }
    return (long long)a;
  }
  long long gcd(long long a, long long b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  long long pw(long long n, long long k) {
    if (k == 0) return 1;
    long long t = pw(n, k / 2);
    long long ans = (t * t) % mod;
    if (k % 2)
      return (ans * n) % mod;
    else
      return ans;
  }
  long long ans = 1e18;
  long long n, k;
  void mn(string s) {
    set<char> st;
    for (auto ch : s) st.insert(ch);
    if (s.length() < n || st.size() > k) return;
    long long num = stoll(s);
    ans = min(ans, num);
  }
  void solve() {
    string s;
    cin >> s;
    cin >> k;
    set<char> st;
    n = s.length();
    ans = 1e18;
    string cur = "";
    for (long long i = 0; i < n; i++) {
      if (st.size() > k) break;
      string nw = cur;
      if (s[i] == '9') {
        if (i == 0) {
          if (k == 1) {
            for (long long i = 0; i < n + 1; i++) nw += '1';
          } else {
            nw += '1';
            for (long long i = 0; i < n; i++) nw += '0';
          }
          mn(nw);
        }
        st.insert('9');
        cur += '9';
        continue;
      } else if (i == n - 1) {
        char ch = s[i];
        if (st.size() < k || st.find(ch) != st.end()) {
          cur += ch;
          mn(cur);
        } else {
          auto it = st.upper_bound(s[i]);
          if (it != st.end()) {
            cur += *it;
            mn(cur);
          }
        }
        break;
      }
      char ch = s[i] + 1;
      if (st.size() == k) {
        auto it = st.lower_bound(ch);
        if (it != st.end()) {
          nw += *it;
          while (nw.length() < s.length()) {
            nw += *st.begin();
          }
          mn(nw);
        }
      } else {
        long long op = 0;
        ch = s[i] + 1;
        if (st.find(ch) == st.end()) {
          op++;
        }
        nw += ch;
        char nxt;
        if (st.find('0') != st.end() || st.size() + op < k) {
          nxt = '0';
        } else {
          nxt = ch;
          if (st.size() > 0) nxt = min(nxt, *st.begin());
        }
        while (nw.length() < s.length()) {
          nw += nxt;
        }
        mn(nw);
      }
      st.insert(s[i]);
      cur += s[i];
    }
    if (cur.length() == s.length()) mn(cur);
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    t = 1;
    cin >> t;
    while (t--) solve();
  }
  