#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
using ll = long long;
const int N = 200005, K = 10, MOD = 1e9 + 7;
int t, n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    string s = to_string(n);
    int len = s.length();
    ll val = 0;
    if (k == 1) {
      int d = s[0] - '0';
      if (d == 9)
        val = 1000000005;
      else {
        ++d;
        for (int i = 0; i < len; ++i) {
          val = val * 10 + d;
        }
      }
    } else {
      int d = s[0] - '0';
      if (d == 9)
        val = 1000000005;
      else {
        ++d;
        for (int i = 0; i < len; ++i) {
          val = val * 10 + (i == 0 ? d : 0);
        }
      }
    }
    for (int i = 0; i < len; ++i) {
      if (i != len - 1 && s[i + 1] == '9') continue;
      ll vl2 = 0;
      vector<char> dv(10, 0);
      for (int j = 0; j <= i; ++j) {
        dv[s[j] - '0'] = 1;
      }
      int cnt = 0;
      for (int j = 0; j < 10; ++j) {
        cnt += dv[j];
      }
      if (cnt > k) break;
      if (i == len - 1) {
        val = n;
        break;
      }
      if (cnt == k) {
        for (int j = 0; j <= i; ++j) {
          vl2 = vl2 * 10 + (s[j] - '0');
        }
        int mn = 0;
        while (!dv[mn]) ++mn;
        int lg = s[i + 1] - '0' + 1;
        while (lg < 10 && !dv[lg]) ++lg;
        if (lg == 10) continue;
        vl2 = vl2 * 10 + lg;
        for (int j = i + 2; j < len; ++j) {
          vl2 = vl2 * 10 + mn;
        }
        val = min(val, vl2);
      } else {
        for (int j = 0; j <= i; ++j) {
          vl2 = vl2 * 10 + (s[j] - '0');
        }
        vl2 = vl2 * 10 + s[i + 1] - '0' + 1;
        int mn = 0;
        if (dv[s[i + 1] - '0' + 1] || cnt < k - 1) {
          mn = 0;
        } else {
          dv[s[i + 1] - '0' + 1] = 1;
          while (!dv[mn]) ++mn;
        }
        for (int j = i + 2; j < len; ++j) {
          vl2 = vl2 * 10 + mn;
        }
        val = min(val, vl2);
      }
    }
    cout << val << '\n';
  }
  return 0;
}

### 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;
using ull = unsigned long long;
using ll = long long;
const int N = 200005, K = 10, MOD = 1e9 + 7;
int t, n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    string s = to_string(n);
    int len = s.length();
    ll val = 0;
    if (k == 1) {
      int d = s[0] - '0';
      if (d == 9)
        val = 1000000005;
      else {
        ++d;
        for (int i = 0; i < len; ++i) {
          val = val * 10 + d;
        }
      }
    } else {
      int d = s[0] - '0';
      if (d == 9)
        val = 1000000005;
      else {
        ++d;
        for (int i = 0; i < len; ++i) {
          val = val * 10 + (i == 0 ? d : 0);
        }
      }
    }
    for (int i = 0; i < len; ++i) {
      if (i != len - 1 && s[i + 1] == '9') continue;
      ll vl2 = 0;
      vector<char> dv(10, 0);
      for (int j = 0; j <= i; ++j) {
        dv[s[j] - '0'] = 1;
      }
      int cnt = 0;
      for (int j = 0; j < 10; ++j) {
        cnt += dv[j];
      }
      if (cnt > k) break;
      if (i == len - 1) {
        val = n;
        break;
      }
      if (cnt == k) {
        for (int j = 0; j <= i; ++j) {
          vl2 = vl2 * 10 + (s[j] - '0');
        }
        int mn = 0;
        while (!dv[mn]) ++mn;
        int lg = s[i + 1] - '0' + 1;
        while (lg < 10 && !dv[lg]) ++lg;
        if (lg == 10) continue;
        vl2 = vl2 * 10 + lg;
        for (int j = i + 2; j < len; ++j) {
          vl2 = vl2 * 10 + mn;
        }
        val = min(val, vl2);
      } else {
        for (int j = 0; j <= i; ++j) {
          vl2 = vl2 * 10 + (s[j] - '0');
        }
        vl2 = vl2 * 10 + s[i + 1] - '0' + 1;
        int mn = 0;
        if (dv[s[i + 1] - '0' + 1] || cnt < k - 1) {
          mn = 0;
        } else {
          dv[s[i + 1] - '0' + 1] = 1;
          while (!dv[mn]) ++mn;
        }
        for (int j = i + 2; j < len; ++j) {
          vl2 = vl2 * 10 + mn;
        }
        val = min(val, vl2);
      }
    }
    cout << val << '\n';
  }
  return 0;
}

```