向中看齐实际上就是向左看齐和向右看齐的结合，以中间的人为分界，左边的向右看，右边的向左看，向右看已经考过了，这也相当于复习，唯一要注意的就是如果人数为奇数，有个处于分界线的人要向左边看的条件，具体见代码。

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
using u64 = unsigned long long;
using u32 = unsigned;
typedef pair<int, int> PII;
const int N = 1010;
int ans[N], a[N];
int n;
void solve() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		ans[i] = a[i];
	}
	int mid = n / 2;
	for (int i = 1; i <= mid; i++)
		for (int j = i + 1; j <= n; j++)
			if (a[j] >= a[i]) {
				ans[i] = a[j];
				break;
			}

	for (int i = mid + 1; i <= n; i++)
		for (int j = i - 1; j >= 1; j--)
			if (a[j] >= a[i]) {
				ans[i] = a[j];
				break;
			}
	for (int i = 1; i <= n; i++)
		cout << ans[i] << ' ';
	cout << '\n';
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin >> t;
	while (t--) {
		solve();
	}

	return 0;
}