#include <bits/stdc++.h>
using namespace std;
#pragma gcc optimize("O3,unroll-loop")
#pragma gcc target( \
    "sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,abm,mmx,popcnt,tune=native")
const int INF = 1111111112;
template <typename T>
void amin(T &a, T b) {
  if (a > b) a = b;
}
bool f(int x, int s) {
  while (s) {
    if (s % 10 == x) return true;
    s /= 10;
  }
  return false;
}
void solve() {
  int n, k;
  cin >> n >> k;
  if (n == 1000000000) {
    if (k == 1)
      cout << 1111111111 << '\n';
    else
      cout << n << '\n';
    return;
  }
  vector<int> v;
  while (n) {
    v.emplace_back(n % 10);
    n /= 10;
  }
  v.emplace_back(0);
  reverse(v.begin(), v.end());
  int m = ((int)v.size()) - 1;
  int dp[11][11][2] = {};
  for (int i = 0; i < (11); ++i)
    for (int j = 0; j < (11); ++j) dp[i][j][0] = dp[i][j][1] = INF;
  dp[0][0][0] = 0;
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= k; ++j) {
      for (int u = 0; u < 10; ++u) {
        if (i == 1 && u == 0) continue;
        if (f(u, dp[i - 1][j][1]) && dp[i - 1][j][1] < INF)
          amin(dp[i][j][1], dp[i - 1][j][1] * 10 + u);
        if (f(u, dp[i - 1][j][0]) && u > v[i] && dp[i - 1][j][0] < INF)
          amin(dp[i][j][1], dp[i - 1][j][0] * 10 + u);
        if (f(u, dp[i - 1][j][0]) && u >= v[i] && dp[i - 1][j][0] < INF)
          amin(dp[i][j][0], dp[i - 1][j][0] * 10 + u);
        if (dp[i - 1][j - 1][1] < INF)
          amin(dp[i][j][1], dp[i - 1][j - 1][1] * 10 + u);
        if (u > v[i] && dp[i - 1][j - 1][0] < INF)
          amin(dp[i][j][1], dp[i - 1][j - 1][0] * 10 + u);
        if (u >= v[i] && dp[i - 1][j - 1][0] < INF)
          amin(dp[i][j][0], dp[i - 1][j - 1][0] * 10 + u);
      }
    }
  }
  int ans = INF;
  for (int i = 0; i <= k; ++i) amin(ans, min(dp[m][i][0], dp[m][i][1]));
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  for (int i = 1; i <= (T); ++i) {
    solve();
  }
}

### Prompt

Your task is to create 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;
#pragma gcc optimize("O3,unroll-loop")
#pragma gcc target( \
    "sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,abm,mmx,popcnt,tune=native")
const int INF = 1111111112;
template <typename T>
void amin(T &a, T b) {
  if (a > b) a = b;
}
bool f(int x, int s) {
  while (s) {
    if (s % 10 == x) return true;
    s /= 10;
  }
  return false;
}
void solve() {
  int n, k;
  cin >> n >> k;
  if (n == 1000000000) {
    if (k == 1)
      cout << 1111111111 << '\n';
    else
      cout << n << '\n';
    return;
  }
  vector<int> v;
  while (n) {
    v.emplace_back(n % 10);
    n /= 10;
  }
  v.emplace_back(0);
  reverse(v.begin(), v.end());
  int m = ((int)v.size()) - 1;
  int dp[11][11][2] = {};
  for (int i = 0; i < (11); ++i)
    for (int j = 0; j < (11); ++j) dp[i][j][0] = dp[i][j][1] = INF;
  dp[0][0][0] = 0;
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= k; ++j) {
      for (int u = 0; u < 10; ++u) {
        if (i == 1 && u == 0) continue;
        if (f(u, dp[i - 1][j][1]) && dp[i - 1][j][1] < INF)
          amin(dp[i][j][1], dp[i - 1][j][1] * 10 + u);
        if (f(u, dp[i - 1][j][0]) && u > v[i] && dp[i - 1][j][0] < INF)
          amin(dp[i][j][1], dp[i - 1][j][0] * 10 + u);
        if (f(u, dp[i - 1][j][0]) && u >= v[i] && dp[i - 1][j][0] < INF)
          amin(dp[i][j][0], dp[i - 1][j][0] * 10 + u);
        if (dp[i - 1][j - 1][1] < INF)
          amin(dp[i][j][1], dp[i - 1][j - 1][1] * 10 + u);
        if (u > v[i] && dp[i - 1][j - 1][0] < INF)
          amin(dp[i][j][1], dp[i - 1][j - 1][0] * 10 + u);
        if (u >= v[i] && dp[i - 1][j - 1][0] < INF)
          amin(dp[i][j][0], dp[i - 1][j - 1][0] * 10 + u);
      }
    }
  }
  int ans = INF;
  for (int i = 0; i <= k; ++i) amin(ans, min(dp[m][i][0], dp[m][i][1]));
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  for (int i = 1; i <= (T); ++i) {
    solve();
  }
}

```