No.1359 [Zelkova 3rd Tune] Four-Person Saison
Notice
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Problem Statement
A tournament was held with $N$ people each from the Spring Country, the Summer Country, the Autumn Country, and the Winter Country, for a total of $4N$ people. As the groups that will compete in the preliminary round of this tournament, we form $N$ groups $G_1,\dots,G_N$, each consisting of one person from each of the Spring, Summer, Autumn, and Winter Countries. In addition, each participant of this tournament has a rating, and using these values, the following are defined.
- Let $X_k,Y_k$ be the maximum and the minimum, respectively, of the ratings of the four people in group $G_k$. The unfairness $D_k$ of group $G_k$ is defined as $D_k:=X_k-Y_k$.
- The unfairness of the preliminary round is defined as $\displaystyle \sum_{k=1}^N D_k^K$.
Find the minimum value $F$ of the unfairness of the preliminary round. Since $F$ can be very large, output $F$ modulo $M$.
Note that you are not asked to find the minimum value of the unfairness of the preliminary round modulo $M$.
Here, the ratings of the $i$-th participants of the Spring, Summer, Autumn, and Winter Countries are $P_i,E_i,A_i,H_i$, respectively.
Constraints
- $1 \leq N \leq 2 \times 10^5$
- $1 \leq K \leq 10^9$
- $1 \leq M \leq 10^9$
- $0 \leq P_i,E_i,A_i,H_i \leq 10^9~(i=1,2,\dots,N)$
- All inputs are integers.
Input
$N\ K\ M$ $P_1\ P_2\ \dots \ P_N$ $E_1\ E_2\ \dots \ E_N$ $A_1\ A_2\ \dots \ A_N$ $H_1\ H_2\ \dots \ H_N$
Output
Output the minimum value of the unfairness of the preliminary round achievable over all possible groupings, modulo $M$. Do not forget the newline.
Sample
Sample 1
Input
2 2 100 1 1000 1002 2 3 1004 1006 4
Output
45
If the two groups are formed so that their ratings are $(1,2,3,4), (1000,1002,1004,1006)$, the unfairness values of the two groups are $3,6$ respectively, so the unfairness of the preliminary round is $3^2+6^2=45$. Moreover, no matter how the groups are formed, the unfairness of the preliminary round cannot be made less than $45$.
Sample 2
Input
1 4 10000 5 10 15 20
Output
625
The only possible grouping is $(5,10,15,20)$, and the unfairness of the preliminary round is $15^4=50625$, so the answer is $625$, the remainder when this is divided by $10000$.
Sample 3
Input
9 20210122 123456789 14 14 21 35 62 37 30 95 4 88 1 68 87 24 20 96 98 7 85 69 67 18 75 37 69 48 7 31 76 67 97 37 99 7 32 47
Output
39598529
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