#include <bits/stdc++.h>
using namespace std;
int dp[10];
int find_greater(int a) {
  int index = -1;
  for (int i = a + 1; i < 10; i++) {
    if (dp[i]) {
      index = i;
      break;
    }
  }
  return index;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    memset(dp, 0, sizeof(dp));
    string ans = "";
    int mnm = 10, mxm = -1;
    vector<int> v;
    int temp = n, cnt = 0;
    while (temp != 0) {
      v.push_back(temp % 10);
      temp /= 10;
      cnt++;
    }
    reverse(v.begin(), v.end());
    int x = 0;
    for (int i = 0; i < cnt; i++) {
      if (x == 1) {
        if (k)
          mnm = 0;
        else {
          mnm = 10;
          for (int j = 0; j < 10; j++) {
            if (dp[j]) {
              mnm = j;
              break;
            }
          }
        }
        ans += char(mnm + '0');
        continue;
      }
      if (dp[v[i]] == 0) {
        if (k == 0) {
          int j;
          int temp = find_greater(v[i]);
          if (temp != -1) {
            ans += char(temp + '0');
            x = 1;
          } else {
            int j = v[i];
            while (i >= 0) {
              i--;
              temp = find_greater(v[i]);
              ans.pop_back();
              dp[v[i]]--;
              if (dp[v[i]] == 0) k++;
              if (temp == -1 && k || (k && temp > v[i] + 1)) {
                k--;
                dp[v[i] + 1] = 1;
                ans += char(v[i] + 1 + '0');
                x = 1;
                break;
              } else if (temp != -1) {
                dp[temp]++;
                ans += char(temp + '0');
                x = 1;
                break;
              }
            }
          }
        } else {
          ans += char(v[i] + '0');
          dp[v[i]] = 1;
          k--;
          mnm = min(mnm, v[i]);
        }
      } else {
        ans += char(v[i] + '0');
        dp[v[i]]++;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Create a solution in cpp for 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 dp[10];
int find_greater(int a) {
  int index = -1;
  for (int i = a + 1; i < 10; i++) {
    if (dp[i]) {
      index = i;
      break;
    }
  }
  return index;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    memset(dp, 0, sizeof(dp));
    string ans = "";
    int mnm = 10, mxm = -1;
    vector<int> v;
    int temp = n, cnt = 0;
    while (temp != 0) {
      v.push_back(temp % 10);
      temp /= 10;
      cnt++;
    }
    reverse(v.begin(), v.end());
    int x = 0;
    for (int i = 0; i < cnt; i++) {
      if (x == 1) {
        if (k)
          mnm = 0;
        else {
          mnm = 10;
          for (int j = 0; j < 10; j++) {
            if (dp[j]) {
              mnm = j;
              break;
            }
          }
        }
        ans += char(mnm + '0');
        continue;
      }
      if (dp[v[i]] == 0) {
        if (k == 0) {
          int j;
          int temp = find_greater(v[i]);
          if (temp != -1) {
            ans += char(temp + '0');
            x = 1;
          } else {
            int j = v[i];
            while (i >= 0) {
              i--;
              temp = find_greater(v[i]);
              ans.pop_back();
              dp[v[i]]--;
              if (dp[v[i]] == 0) k++;
              if (temp == -1 && k || (k && temp > v[i] + 1)) {
                k--;
                dp[v[i] + 1] = 1;
                ans += char(v[i] + 1 + '0');
                x = 1;
                break;
              } else if (temp != -1) {
                dp[temp]++;
                ans += char(temp + '0');
                x = 1;
                break;
              }
            }
          }
        } else {
          ans += char(v[i] + '0');
          dp[v[i]] = 1;
          k--;
          mnm = min(mnm, v[i]);
        }
      } else {
        ans += char(v[i] + '0');
        dp[v[i]]++;
      }
    }
    cout << ans << endl;
  }
}

```