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: Accepted)

fun main(args: Array<String>) {
    val (n, k) = readLine()?.split(" ")?.map(String::toLong) ?: return
    val a = readLine()?.split(" ")?.map(String::toInt) ?: return
    val m = (1e9.toLong() + 7L)
    var t = 0L
    val k1 = ((k + 1) * ((k) / 2) + if (((k) % 2L) == 0L) {
        0
    } else {
        (k + 1) / 2
    }) % m
    val k2 = ((k) * ((k - 1) / 2) + if (((k - 1) % 2L) == 0L) {
        0
    } else {
        k / 2
    }) % m

    0.until(n.toInt()).forEach { i ->
        var f = 0L
        var s = 0L
        0.until(n.toInt()).forEach { j ->
            if (a[i] > a[j]) {
                if (i > j) {
                    s++
                } else {
                    f++
                }
            }
        }
        t = (t + (f * k1)) % m
        t = (t + (s * k2)) % m
    }
    println(t.toString())
}