#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
int digit[maxn];
int dp[10][2][1 << 10];
int n, k;
int g[1 << 10];
int dfs(int len, bool lim, int s) {
  if (len == 0) return g[s] <= k;
  int &x = dp[len][lim][s];
  if (~x) return x;
  int low = lim ? digit[len] : 0;
  int ans = false;
  for (int i = low; i <= 9; i++) {
    ans |= dfs(len - 1, lim && i == low, s | (1 << i));
    if (ans) break;
  }
  return x = ans;
}
long long ksm(long long a, long long b) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) ans = ans * base % mod;
    b >>= 1;
    base = base * base % mod;
  }
  return ans;
}
long long getans(int len, int lim, int s) {
  if (len == 0) return 0;
  int low = lim ? digit[len] : 0;
  for (int i = low; i <= 9; i++) {
    bool ok = dfs(len - 1, lim && i == low, s | (1 << i));
    if (ok) {
      return i * ksm(10, len - 1) +
             getans(len - 1, lim && i == low, s | (1 << i));
    }
  }
  return -1;
}
long long solve() {
  int k = 0;
  while (n) {
    digit[++k] = n % 10;
    n /= 10;
  }
  memset(dp, -1, sizeof dp);
  dfs(k, true, 0);
  return getans(k, true, 0);
}
int main() {
  ios::sync_with_stdio(false);
  g[0] = 0;
  for (int i = 1; i < 1 << 10; i++) {
    g[i] = g[i >> 1] + (i & 1);
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << solve() << endl;
  }
  return 0;
}

### Prompt

Generate 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;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
int digit[maxn];
int dp[10][2][1 << 10];
int n, k;
int g[1 << 10];
int dfs(int len, bool lim, int s) {
  if (len == 0) return g[s] <= k;
  int &x = dp[len][lim][s];
  if (~x) return x;
  int low = lim ? digit[len] : 0;
  int ans = false;
  for (int i = low; i <= 9; i++) {
    ans |= dfs(len - 1, lim && i == low, s | (1 << i));
    if (ans) break;
  }
  return x = ans;
}
long long ksm(long long a, long long b) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) ans = ans * base % mod;
    b >>= 1;
    base = base * base % mod;
  }
  return ans;
}
long long getans(int len, int lim, int s) {
  if (len == 0) return 0;
  int low = lim ? digit[len] : 0;
  for (int i = low; i <= 9; i++) {
    bool ok = dfs(len - 1, lim && i == low, s | (1 << i));
    if (ok) {
      return i * ksm(10, len - 1) +
             getans(len - 1, lim && i == low, s | (1 << i));
    }
  }
  return -1;
}
long long solve() {
  int k = 0;
  while (n) {
    digit[++k] = n % 10;
    n /= 10;
  }
  memset(dp, -1, sizeof dp);
  dfs(k, true, 0);
  return getans(k, true, 0);
}
int main() {
  ios::sync_with_stdio(false);
  g[0] = 0;
  for (int i = 1; i < 1 << 10; i++) {
    g[i] = g[i >> 1] + (i & 1);
  }
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << solve() << endl;
  }
  return 0;
}

```