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: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

const mod = 1000000007

func main() {
	var n, k int
	fmt.Scan(&n, &k)

	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	a := make([]int, n)
	for i := range a {
		sc.Scan()
		a[i], _ = strconv.Atoi(sc.Text())
	}

	ac, cc := 0, 0
	for i := range a {
		for j := range a {
			if a[i] <= a[j] {
				continue
			}
			ac++
			if j >= i {
				cc++
			}
		}
	}
	ans := (k + 2) * (k - 1) / 2 % mod
	ans = (ans*ac + cc) % mod
	fmt.Println(ans)

}