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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    val (N, K) = readLine()!!.split(" ").map { it.toLong() }
    val A = readLine()!!.split(" ").map { it.toLong() }.toLongArray()
    
    var n = N.toInt()


    val all = LongArray(n, { 0 })
    val onlyAfter = LongArray(n, { 0 })

    val MOD = 1000000000 + 7
    var answer = 0L

    (0 until n).forEach { i ->
        (0 until n).forEach { j ->
            if (A[i] > A[j]) {
                if (j >= i) {
                    onlyAfter[i]++
                }
                all[i]++
            }
        }
    }

    (0 until n).forEach { i ->
        var local = 0L
        // For K
        local = (local + onlyAfter[i] * K) % MOD
        local = (local + (K * ((all[i] * (K-1))) / 2)) % MOD
        answer = (answer + local).mod(MOD)
    }

    println(answer)
}