#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  debug& operator<<(c x) {
    cerr << x;
    return *this;
  }
  template <class c, class d>
  debug& operator<<(pair<c, d> x) {
    *this << "(" << x.first << ", " << x.second << ")";
    return *this;
  }
  template <class c>
  debug& operator<<(vector<c> x) {
    *this << "[";
    int f = 0;
    for (auto it : x) {
      *this << (f ? ", " : "");
      *this << it;
      f = 1;
    }
    *this << "]";
    return *this;
  }
};
bool f(int pos, string cur, set<int> digits, bool smaller, int k, int len,
       string n) {
  if ((int)digits.size() > k) {
    return false;
  }
  if (pos == len) {
    printf("%s\n", cur.c_str());
    return true;
  }
  for (char dig = '0'; dig <= '9'; dig++) {
    bool flag = smaller;
    if (smaller) {
    } else {
      if (dig < n[pos]) continue;
      if (dig == n[pos]) {
      } else if (dig > n[pos]) {
        flag = true;
      }
    }
    int already = (digits.find(dig - '0') != digits.end());
    digits.insert(dig - '0');
    cur += dig;
    int done = f(pos + 1, cur, digits, flag, k, len, n);
    if (done) {
      return true;
    }
    cur.pop_back();
    if (!already) {
      digits.erase(dig - '0');
    }
  }
  return false;
}
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  string t = "";
  set<int> s;
  f(0, t, s, 0, k, (int)to_string(n).length(), to_string(n));
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  debug& operator<<(c x) {
    cerr << x;
    return *this;
  }
  template <class c, class d>
  debug& operator<<(pair<c, d> x) {
    *this << "(" << x.first << ", " << x.second << ")";
    return *this;
  }
  template <class c>
  debug& operator<<(vector<c> x) {
    *this << "[";
    int f = 0;
    for (auto it : x) {
      *this << (f ? ", " : "");
      *this << it;
      f = 1;
    }
    *this << "]";
    return *this;
  }
};
bool f(int pos, string cur, set<int> digits, bool smaller, int k, int len,
       string n) {
  if ((int)digits.size() > k) {
    return false;
  }
  if (pos == len) {
    printf("%s\n", cur.c_str());
    return true;
  }
  for (char dig = '0'; dig <= '9'; dig++) {
    bool flag = smaller;
    if (smaller) {
    } else {
      if (dig < n[pos]) continue;
      if (dig == n[pos]) {
      } else if (dig > n[pos]) {
        flag = true;
      }
    }
    int already = (digits.find(dig - '0') != digits.end());
    digits.insert(dig - '0');
    cur += dig;
    int done = f(pos + 1, cur, digits, flag, k, len, n);
    if (done) {
      return true;
    }
    cur.pop_back();
    if (!already) {
      digits.erase(dig - '0');
    }
  }
  return false;
}
void solve() {
  int n, k;
  scanf("%d%d", &n, &k);
  string t = "";
  set<int> s;
  f(0, t, s, 0, k, (int)to_string(n).length(), to_string(n));
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
}

```