#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int k;
    cin >> k;
    vector<int> a(10, 0);
    int k1 = k;
    int pos = s.size();
    for (int i = 0; i < s.size(); ++i) {
      a[s[i] - '0']++;
      if (a[s[i] - '0'] == 1) {
        --k1;
      }
      if (k1 == -1) {
        a[s[i] - '0']--;
        pos = i;
        break;
      }
    }
    if (pos == s.size()) {
      cout << s << '\n';
      continue;
    }
    int ch = -1;
    char smth;
    int zero = 0;
    bool ok = 0;
    for (int i = s[pos] - '0' + 1; i <= 9; ++i) {
      if (a[i] > 0) {
        ch = pos;
        smth = i + '0';
        ok = 1;
        break;
      }
    }
    if (!ok) {
      for (int i = pos - 1; i >= 0; --i) {
        if (a[s[i] - '0'] == 1 && s[i] != '9') {
          ch = i;
          a[s[i] - '0' + 1]++;
          smth = s[i] + 1;
          ok = 1;
        } else
          for (int j = s[i] - '0' + 1; j <= 9; ++j) {
            if (a[j] > 0) {
              ch = i;
              smth = j + '0';
              ok = 1;
              break;
            }
          }
        a[s[i] - '0']--;
        if (ok) break;
      }
    }
    if (ch != -1) {
      char minn;
      s[ch] = smth;
      int cnt = 0;
      for (int i = 0; i < 10; ++i) {
        if (a[i] > 0) ++cnt;
      }
      for (int i = 0; i < 10; ++i) {
        if (a[i] > 0) {
          minn = i + '0';
          break;
        }
      }
      if (cnt < k) minn = '0';
      for (int i = ch + 1; i < s.size(); ++i) {
        s[i] = minn;
      }
      cout << s << '\n';
    }
  }
}

### 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int k;
    cin >> k;
    vector<int> a(10, 0);
    int k1 = k;
    int pos = s.size();
    for (int i = 0; i < s.size(); ++i) {
      a[s[i] - '0']++;
      if (a[s[i] - '0'] == 1) {
        --k1;
      }
      if (k1 == -1) {
        a[s[i] - '0']--;
        pos = i;
        break;
      }
    }
    if (pos == s.size()) {
      cout << s << '\n';
      continue;
    }
    int ch = -1;
    char smth;
    int zero = 0;
    bool ok = 0;
    for (int i = s[pos] - '0' + 1; i <= 9; ++i) {
      if (a[i] > 0) {
        ch = pos;
        smth = i + '0';
        ok = 1;
        break;
      }
    }
    if (!ok) {
      for (int i = pos - 1; i >= 0; --i) {
        if (a[s[i] - '0'] == 1 && s[i] != '9') {
          ch = i;
          a[s[i] - '0' + 1]++;
          smth = s[i] + 1;
          ok = 1;
        } else
          for (int j = s[i] - '0' + 1; j <= 9; ++j) {
            if (a[j] > 0) {
              ch = i;
              smth = j + '0';
              ok = 1;
              break;
            }
          }
        a[s[i] - '0']--;
        if (ok) break;
      }
    }
    if (ch != -1) {
      char minn;
      s[ch] = smth;
      int cnt = 0;
      for (int i = 0; i < 10; ++i) {
        if (a[i] > 0) ++cnt;
      }
      for (int i = 0; i < 10; ++i) {
        if (a[i] > 0) {
          minn = i + '0';
          break;
        }
      }
      if (cnt < k) minn = '0';
      for (int i = ch + 1; i < s.size(); ++i) {
        s[i] = minn;
      }
      cout << s << '\n';
    }
  }
}

```