题意：

q次询问，每次询问给你长度为n的排列，然后你每次可以选择一个位置i和i+1的数字进行交换。但是每个位置只能交换一次，问你反转若干次后，这个排列最小是多少？

题目：

You are given a permutation of length n. Recall that the permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

You can perform at most n−1 operations with the given permutation (it is possible that you don’t perform any operations at all). The i-th operation allows you to swap elements of the given permutation on positions i and i+1. Each operation can be performed at most once. The operations can be performed in arbitrary order.

Your task is to find the lexicographically minimum possible permutation obtained by performing some of the given operations in some order.

You can see the definition of the lexicographical order in the notes section.

You have to answer q independent test cases.

For example, let’s consider the permutation [5,4,1,3,2]. The minimum possible permutation we can obtain is [1,5,2,4,3] and we can do it in the following way:

perform the second operation (swap the second and the third elements) and obtain the permutation [5,1,4,3,2];
perform the fourth operation (swap the fourth and the fifth elements) and obtain the permutation [5,1,4,2,3];
perform the third operation (swap the third and the fourth elements) and obtain the permutation [5,1,2,4,3].
perform the first operation (swap the first and the second elements) and obtain the permutation [1,5,2,4,3];
Another example is [1,2,4,3]. The minimum possible permutation we can obtain is [1,2,3,4] by performing the third operation (swap the third and the fourth elements).

Input

The first line of the input contains one integer q (1≤q≤100) — the number of test cases. Then q test cases follow.

The first line of the test case contains one integer n (1≤n≤100) — the number of elements in the permutation.

The second line of the test case contains n distinct integers from 1 to n — the given permutation.

Output

For each test case, print the answer on it — the lexicograhically minimum possible permutation obtained by performing some of the given operations in some order.

Example

Input

4
5
5 4 1 3 2
4
1 2 4 3
1
1
4
4 3 2 1

Output

1 5 2 4 3
1 2 3 4
1
1 4 3 2

Note

Recall that the permutation p of length n is lexicographically less than the permutation q of length n if there is such index i≤n that for all j from 1 to i−1 the condition pj=qj is satisfied, and pi<qi. For example:

p=[1,3,5,2,4] is less than q=[1,3,5,4,2] (such i=4 exists, that pi<qi and for each j<i holds pj=qj),
p=[1,2] is less than q=[2,1] (such i=1 exists, that pi<qi and for each j<i holds pj=qj).

官方题解：

The following greedy solution works: let’s take the minimum element and move it to the leftmost position we can. With this algorithm, all forbidden operations are form the prefix of operations: (1,2), (2,3), …, and so on. So we can carry the position of the leftmost operation we can perform pos. Initially, it is 1. We repeat the algorithm until pos≥n. Let’s find the position of the minimum element among elements apos,apos+1,…,an. Let this position be nxt. If nxt=pos then let’s increase pos and continue the algorithm. Otherwise, we need to move the element from the position nxt to the position pos and then set pos:=nxt.

Time complexity: O(n2).

谷歌翻译：

以下贪婪的解决方案有效：让我们取最小元素并将其移到我们可以的最左边位置。 使用此算法，所有禁止的操作都是操作的前缀：（1,2），（2,3），…等。 因此，我们可以携带可以执行pos的最左侧操作的位置。 最初为1。我们重复该算法，直到pos≥n。 让我们找到最小的元素在元素’，’+ 1，…，an中的位置。 将此位置设为nxt。 如果nxt = pos，那么让我们增加pos并继续算法。 否则，我们需要将元素从位置nxt移到位置pos，然后设置pos：= nxt。

思路：

因为每个位子只能一次操作，即每次找到当前最小数的位置，交换即可；用贪心

AC代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int t,n,dp[110],book[110];
int main()
{scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));memset(book,0,sizeof(book));scanf("%d",&n);for(int i=1; i<=n; i++)scanf("%d",&dp[i]);int k=1,w=n-1;while(w>0&&k<=n){for(int i=1; i<=n; i++)if(dp[i]==k){for(int j=i-1; j>=k; j--){if(book[j])continue;if(dp[j]<dp[j+1])break;swap(dp[j],dp[j+1]);book[j]=1;w--;}k++;break;}}for(int i=1; i<=n; i++)printf("%d%c",dp[i],i==n?'\n':' ');}return 0;
}