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)

import java.util.*

fun main(args: Array<String>) {
	val mod = 1000000007

	val scanner = Scanner(System.`in`)
	val n = scanner.nextInt()
	val k = scanner.nextLong()

	val map = IntArray(2001){i -> 0}

	var cnt = 0L
	var max = 0
	for (i in 0..n-1){
		val a = scanner.nextInt()
		for (j in a+1..2000) cnt+=map[j]
		map[a]++
		max = Math.max(max, a)
	}
	cnt = (cnt*k)%mod
	var tmp = 0L
	for (i in max downTo 1){
		cnt = (cnt+tmp*map[i]*(((k-1)*k/2)%mod))%mod
		tmp += map[i]
	}

	println(cnt)
}