#include <bits/stdc++.h>
using namespace std;
void SolveF();
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  SolveF();
  return 0;
}
void SolveF() {
  int T;
  cin >> T;
  while (T--) {
    long long n;
    int k;
    cin >> n >> k;
    int currk[11];
    int minInt[11];
    int a[10];
    memset(a, 0, sizeof(a));
    string s = to_string(n);
    int lens = s.length();
    ++a[s[0] - '0'];
    currk[0] = 1;
    minInt[0] = s[0] - '0';
    for (int i = 1; i < lens; ++i) {
      if (a[s[i] - '0'] == 0) {
        currk[i] = currk[i - 1] + 1;
        minInt[i] =
            ((minInt[i - 1]) < (s[i] - '0') ? (minInt[i - 1]) : (s[i] - '0'));
      } else {
        currk[i] = currk[i - 1];
        minInt[i] = minInt[i - 1];
      }
      ++a[s[i] - '0'];
    }
    if (k == currk[lens - 1]) {
      cout << s << endl;
      continue;
    } else if (k > lens) {
      cout << s << endl;
    } else if (k > currk[lens - 1]) {
      cout << s << endl;
    } else {
      while (1) {
        int pos = 0;
        for (int i = 0; i < 10; ++i) {
          a[i] = 0;
          currk[i] = 0;
        }
        currk[0] = 1;
        a[s[0] - '0'] = 1;
        for (int i = 1; i < lens; ++i) {
          if (a[s[i] - '0'] == 0)
            currk[i] = currk[i - 1] + 1;
          else
            currk[i] = currk[i - 1];
          ++a[s[i] - '0'];
        }
        if (k >= currk[lens - 1]) break;
        for (int i = 0; i < lens; ++i) {
          if (currk[i] == k + 1) {
            pos = i;
            break;
          }
        }
        while (s[pos] == 9 + '0') --pos;
        ++s[pos];
        for (int i = pos + 1; i < lens; ++i) s[i] = '0';
      }
      cout << s << endl;
    }
  }
}

### Prompt

Please formulate 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;
void SolveF();
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  SolveF();
  return 0;
}
void SolveF() {
  int T;
  cin >> T;
  while (T--) {
    long long n;
    int k;
    cin >> n >> k;
    int currk[11];
    int minInt[11];
    int a[10];
    memset(a, 0, sizeof(a));
    string s = to_string(n);
    int lens = s.length();
    ++a[s[0] - '0'];
    currk[0] = 1;
    minInt[0] = s[0] - '0';
    for (int i = 1; i < lens; ++i) {
      if (a[s[i] - '0'] == 0) {
        currk[i] = currk[i - 1] + 1;
        minInt[i] =
            ((minInt[i - 1]) < (s[i] - '0') ? (minInt[i - 1]) : (s[i] - '0'));
      } else {
        currk[i] = currk[i - 1];
        minInt[i] = minInt[i - 1];
      }
      ++a[s[i] - '0'];
    }
    if (k == currk[lens - 1]) {
      cout << s << endl;
      continue;
    } else if (k > lens) {
      cout << s << endl;
    } else if (k > currk[lens - 1]) {
      cout << s << endl;
    } else {
      while (1) {
        int pos = 0;
        for (int i = 0; i < 10; ++i) {
          a[i] = 0;
          currk[i] = 0;
        }
        currk[0] = 1;
        a[s[0] - '0'] = 1;
        for (int i = 1; i < lens; ++i) {
          if (a[s[i] - '0'] == 0)
            currk[i] = currk[i - 1] + 1;
          else
            currk[i] = currk[i - 1];
          ++a[s[i] - '0'];
        }
        if (k >= currk[lens - 1]) break;
        for (int i = 0; i < lens; ++i) {
          if (currk[i] == k + 1) {
            pos = i;
            break;
          }
        }
        while (s[pos] == 9 + '0') --pos;
        ++s[pos];
        for (int i = pos + 1; i < lens; ++i) s[i] = '0';
      }
      cout << s << endl;
    }
  }
}

```