Codeforces Round 908 (Div. 2)

D. Neutral Tonality

https://codeforces.com/contest/1894/problem/D

3 seconds / 512 megabytes

standard input / standard output

Problem

You are given an array a consisting of n integers, as well as an array b consisting of m integers.

Let LIS(c) denote the length of the longest increasing subsequence of array c. For example, LIS([2,1―,1,3―]) = 2, LIS([1―,7―,9―]) = 3, LIS([3,1―,2―,4―]) = 3.

You need to insert the numbers b1,b2,…,bm into the array a, at any positions, in any order. Let the resulting array be c1,c2,…,cn+m. You need to choose the positions for insertion in order to minimize LIS(c).

Formally, you need to find an array c1,c2,…,cn+m that simultaneously satisfies the following conditions:

  • The array a1,a2,…,an is a subsequence of the array c1,c2,…,cn+m.
  • The array c1,c2,…,cn+m consists of the numbers a1,a2,…,an,b1,b2,…,bm, possibly rearranged.
  • The value of LIS(c) is the minimum possible among all suitable arrays c.

Input

Each test contains multiple test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers n,m (1≤n≤2⋅105,1≤m≤2⋅105) — the length of array a and the length of array b.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109) — the elements of the array a.

The third line of each test case contains m integers b1,b2,…,bm (1≤bi≤109) — the elements of the array b.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105, and the sum of m over all test cases does not exceed 2⋅105.

Output

For each test case, output n+m numbers — the elements of the final array c1,c2,…,cn+m, obtained after the insertion, such that the value of LIS(c) is minimized. If there are several answers, you can output any of them.

Example

Input

7
2 1
6 4
5
5 5
1 7 2 4 5
5 4 1 2 7
1 9
7
1 2 3 4 5 6 7 8 9
3 2
1 3 5
2 4
10 5
1 9 2 3 8 1 4 7 2 9
7 8 5 4 6
2 1
2 2
1
6 1
1 1 1 1 1 1
777

Output

6 5 4
1 1 7 7 2 2 4 4 5 5
9 8 7 7 6 5 4 3 2 1
1 3 5 2 4
1 9 2 3 8 8 1 4 4 7 7 2 9 6 5
2 2 1
777 1 1 1 1 1 1

Note

In the first test case, LIS(a)=LIS([6,4])=1. We can insert the number 5 between 6 and 4, then LIS(c)=LIS([6,5,4])=1.

In the second test case, LIS([1―,7,2―,4―,5―]) = 4. After the insertion, c=[1,1,7,7,2,2,4,4,5,5]. It is easy to see that LIS(c)=4. It can be shown that it is impossible to achieve LIS(c) less than 4.

题解

首先需要注意到，将 b 序列插入 a 序列是不可能降低 a 序列原有的 LIS 长度的。题目让我们最小化 LIS 长度，则意义是让我们最小化 ΔLIS.

最基本的，我们插入的序列必须是降序的，如此能保证插入的 b 序列自身不会产生上升子序列，避免 LIS 长度增长。所以首先将 b 序列排序为降序。

接下来，若插入的值与原序列形成的都是降序段，则不会使 LIS 长度增长。因此，可以将每个数都插入到原序列比它小的数左边。对于一个序列 3,1,5，如果要插入 6,4,2，变成 6―,4―,3,2―,1,5，不会导致 LIS 长度增长。

因此，这道题的最优解实际上 ΔLIS=0.

代码实现上其实非常简单，和归并排序的合并过程是一样的写法。

代码

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
 
using namespace std;
 
void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(n), b(m);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    for (int i = 0; i < m; i++)
        cin >> b[i];
    sort(b.begin(), b.end(), greater<int>());
    int i = 0, j = 0;
    while (i < n && j < m)
    {
        if (a[i] <= b[j])
            cout << b[j++] << ' ';
        else
            cout << a[i++] << ' ';
    }
    for (; i < n; i++)
        cout << a[i] << ' ';
    for (; j < m; j++)
        cout << b[j] << ' ';
    cout << endl;
}
 
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    return 0;
}