#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkMax(T &first, T second) {
  if (second > first) first = second;
}
template <typename T>
void chkMin(T &first, T second) {
  if (second < first) first = second;
}
template <typename T>
inline void read(T &f) {
  f = 0;
  T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') {
      fu = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
const int MAX_N = 2e5 + 10, MAX_M = 1e6 + 10;
bool ok(string first, int k) {
  sort(first.begin(), first.end());
  int d = unique(first.begin(), first.end()) - first.begin();
  return d <= k;
}
void solve() {
  string n;
  int k;
  cin >> n >> k;
  if (ok(n, k)) {
    cout << n << '\n';
    return;
  }
  for (int i = n.size() - 1; i >= 0; i--) {
    for (n[i]++; n[i] <= '9'; n[i]++) {
      for (char j = '0'; j <= '9'; j++) {
        for (int k = i + 1; k < n.size(); k++) {
          n[k] = j;
        }
        if (ok(n, k)) {
          cout << n << '\n';
          goto nex;
        }
      }
    }
  }
nex:;
  return;
}
int main() {
  int T = 1;
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
template <typename T>
void chkMax(T &first, T second) {
  if (second > first) first = second;
}
template <typename T>
void chkMin(T &first, T second) {
  if (second < first) first = second;
}
template <typename T>
inline void read(T &f) {
  f = 0;
  T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') {
      fu = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
const int MAX_N = 2e5 + 10, MAX_M = 1e6 + 10;
bool ok(string first, int k) {
  sort(first.begin(), first.end());
  int d = unique(first.begin(), first.end()) - first.begin();
  return d <= k;
}
void solve() {
  string n;
  int k;
  cin >> n >> k;
  if (ok(n, k)) {
    cout << n << '\n';
    return;
  }
  for (int i = n.size() - 1; i >= 0; i--) {
    for (n[i]++; n[i] <= '9'; n[i]++) {
      for (char j = '0'; j <= '9'; j++) {
        for (int k = i + 1; k < n.size(); k++) {
          n[k] = j;
        }
        if (ok(n, k)) {
          cout << n << '\n';
          goto nex;
        }
      }
    }
  }
nex:;
  return;
}
int main() {
  int T = 1;
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}

```