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 Go (1.6) (judge verdict: Accepted)

// https://atcoder.jp/contests/jsc2019-qual/tasks/jsc2019_qual_b

package main

import "fmt"
import "bufio"
import "os"
import "strconv"

var _sw = bufio.NewScanner(os.Stdin)

func main() {
	N := nextInt()
	K := nextInt()
	A := make([]int, N)

	for i := range A {
		A[i] = nextInt()
	}

	w := 0
	h := 0
	for i := range A {
		for j := range A {
			if A[i] > A[j] {
				w++
				if i < j {
					h++
				}
			}
		}
	}

	D := 1000000000 + 7
	ans := (K*h%D + ((K-1)*K/2%D)*w) % D
	fmt.Println(ans)
}

func init() {
	_sw.Split(bufio.ScanWords)
}

func nextInt() int {
	if _sw.Scan() == false {
		panic(_sw.Err())
	}
	i, e := strconv.Atoi(_sw.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func nextString() string {
	if _sw.Scan() == false {
		panic(_sw.Err())
	}
	return _sw.Text()
}

func max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

func min(x, y int) int {
	if x > y {
		return y
	}
	return x
}