Codeforces Global Round 19 |
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You are given a permutation $$$p$$$ of length $$$n$$$.
You can choose any subsequence, remove it from the permutation, and insert it at the beginning of the permutation keeping the same order.
For every $$$k$$$ from $$$0$$$ to $$$n$$$, find the minimal possible number of inversions in the permutation after you choose a subsequence of length exactly $$$k$$$.
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 50\,000$$$) — the number of test cases.
The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 5 \cdot 10^5$$$) — the length of the permutation.
The second line of each test case contains the permutation $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$).
It is guaranteed that the total sum of $$$n$$$ doesn't exceed $$$5 \cdot 10^5$$$.
For each test case output $$$n + 1$$$ integers. The $$$i$$$-th of them must be the answer for the subsequence length of $$$i - 1$$$.
31144 2 1 355 1 3 2 4
0 0 4 2 2 1 4 5 4 2 2 1 5
In the second test case:
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