Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String args[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        long k = Long.parseLong(sc.next());
        Integer[] a = new Integer[n];
        Arrays.setAll(a, i -> Integer.parseInt(sc.next()));
        sc.close();

        // 主処理
        long count = 0;
        long count2 = 0;
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (a[i] < a[j]) {
                    count++; // 転倒数
                    count2++;
                }
            }
            for (int j = 0; j < n; j++) {
                if (a[i] < a[j]) {
                    count2++; // 2周目の転倒数
                }
            }
        }
        long diff = count2 - count; // 転倒数の増分

        final long MOD = 1000000007;
        long result = 0;
        for (int i = 0; i < k; i++) {
            result += count + diff * i;
            result %= MOD;
        }

        // 出力
        System.out.println(result);
    }
}