G. Restore the Permutation

G. Restore the Permutation

A sequence of $n$ numbers is called permutation if it contains all numbers from $1$ to $n$ exactly once. For example, the sequences $[3,1,4,2]$, $[1]$ and $[2,1]$ are permutations, but $[1,2,1]$, $[0,1]$ and $[1,3,4]$ — are not.

For a permutation $p$ of even length $n$ you can make an array $b$ of length $\frac{n}{2}$ such that:

• $b_i = \max(p_{2i - 1}, p_{2i})$ for $1 \le i \le \frac{n}{2}$

For example, if $p = [2,4,3,1,5,6]$, then:

• $b_1 = \max(p_1, p_2) = \max(2, 4) = 4$
• $b_2 = \max(p_3, p_4) = \max(3,1)=3$
• $b_3 = \max(p_5, p_6) = \max(5,6) = 6$

As a result, we made $b = [4,3,6]$.
For a given array $b$, find the lexicographically minimal permutation $p$ such that you can make the given array $b$ from it.

If $b = [4,3,6]$, then the lexicographically minimal permutation from which it can be made is $p = [1,4,2,3,5,6]$, since:

• $b_1 = \max(p_1, p_2) = \max(1, 4) = 4$
• $b_2 = \max(p_3, p_4) = \max(2, 3) = 3$
• $b_3 = \max(p_5, p_6) = \max(5, 6) = 6$

A permutation $x_1, x_2, \ldots, x_n$ is lexicographically smaller than a permutation $x_1, x_2, \ldots, x_n$ if and only if there exists such $i$ $(1 \leq i \leq n)$ that $x_1=y_1,x_2=y_2, \ldots ,x_{i−1}=y_{i−1}$ and $x_i \le y_i$.

Input

The first line of input data contains a single integer $t$ $(1 \leq t \leq {10}^{4})$ — the number of test cases.

The description of the test cases follows.

The first line of each test case contains one even integer $n$ $(2 \leq n \leq 2 \cdot {10}^{5})$.

The second line of each test case contains exactly $\frac{n}{2}$ integers $b_i$ $(1 \leq b_i \leq n)$ — elements of array $b$.

It is guaranteed that the sum of $n$ values over all test cases does not exceed $2 \cdot {10}^{5}$.

Output

For each test case, print on a separate line:

lexicographically minimal permutation $p$ such that you can make an array $b$ from it;
or a number $-1$ if the permutation you are looking for does not exist.

Example

input

6
6
4 3 6
4
2 4
8
8 7 2 3
6
6 4 2
4
4 4
8
8 7 4 5

output

1 4 2 3 5 6 
1 2 3 4 
-1
5 6 3 4 1 2 
-1
1 8 6 7 2 4 3 5 

Note

The first test case is parsed in the problem statement.

 

解题思路

　　这题容易想到先开个$\text{std::set}$存没出现过的数，然后从前往后枚举数组$b$，在集合中找到小于$b_i$的最小数。但这种做法会存在一个问题，因为每次都是选出最小的数，这导致集合中剩下的数都比较大，这就会导致数组$b$往后的数可能会出现无解的情况，比如现在有数组$b = [6, 4, 2]$，如果按照上面的做法来选择，那么对于$6$则选择$1$，$4$则选择$3$，当枚举到$2$时此时集合中只剩下$5$，而$\max \{ {2, 5} \} = 5$就无解了，但实际上最优解是$[5, 6, 3, 4, 1, 2]$。

　　根据贪心的思想，对于数组$b$前面的数我们应尽可能匹配小的数，而后面应尽可能匹配大的数，因此为了尽可能保证优解，我们可以从后面往前枚举数组$b$，在集合中找到小于$b_i$的最大数，如果存在的话那么就选择这个数并从集合中删除，否则说明无解。这种做法就为数组$b$前面的数的选择保留了尽可能小的数，一方面尽可能保证字典序最小，另一方面也尽可能保证前面部分能够有解。

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 10;

int a[N], ans[N];

void solve() {
    int n;
    scanf("%d", &n);
    set<int> st;
    for (int i = 1; i <= n; i++) {
        st.insert(i);
    }
    for (int i = 1; i <= n >> 1; i++) {
        scanf("%d", a + i);
        st.erase(a[i]);
    }
    
    if (st.size() != n >> 1) {    // 说明给定的序列有重复数字出现，一定无解
        printf("-1\n");
        return;
    }
    
    for (int i = n >> 1; i; i--) {
        ans[i * 2] = a[i];
        auto it = st.lower_bound(a[i]);    // 先找到大于等于a[i]的最小的数
        if (it == st.begin()) {    // 如果这个数是集合中最小的数，意味着不存在严格小于a[i]的数，无解
            printf("-1\n");
            return;
        }
        ans[i * 2 - 1] = *--it;    // prev(it)就是小于a[i]的最大的数
        st.erase(it);
    }
    
    for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
    }
    printf("\n");
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #834 (Div. 3) A-G：https://www.cnblogs.com/BlankYang/p/16906184.html