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"
	"sort"
	"strconv"
)

const (
	initialBufSize = 1e4
	maxBufSize     = 1e6 + 7
	mod            = 1e9 + 7
)

var buf []byte = make([]byte, maxBufSize)
var sc = bufio.NewScanner(os.Stdin)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buf, maxBufSize)
}

func main() {
	N, K := nextInt(), nextInt()
	ans := 0
	A := make([]int, N)
	// revnums := make([]int, N)
	revnums1st := 0
	revnums2nd := 0
	// K = (K + (K+1)/2) % mod
	for i := range A {
		A[i] = nextInt()
	}
	for i := 0; i < N; i++ {
		for j := i + 1; j < N; j++ {
			if A[i] > A[j] {
				revnums1st++
			}
		}
	}
	revnums2nd += revnums1st
	for i := 0; i < N; i++ {
		for j := 0; j < N; j++ {
			if A[i] > A[j] {
				revnums2nd++
			}
		}
	}
	d := revnums2nd - revnums1st
	l := revnums1st + (K-1)*d
	// fmt.Println(revnums1st, revnsums2nd, d, l)
	ans = ((revnums1st + l) % mod * K % mod / 2)
	fmt.Println(ans % mod)
}

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

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

func abs(a int) int {
	if a > 0 {
		return a
	} else {
		return -a
	}
}

func pow(p, q int) int {
	res := p
	if q == 0 {
		return 1
	}
	for i := 0; i < q-1; i++ {
		res *= p
	}
	return res
}

func MinOf(vars ...int) int {
	min := vars[0]
	for _, i := range vars {
		if min > i {
			min = i
		}
	}
	return min
}

func MaxOf(vars ...int) int {
	max := vars[0]
	for _, i := range vars {
		if max < i {
			max = i
		}
	}
	return max
}

func gcdof2numbers(a int, b int) int {
	if b == 0 {
		return a
	}
	return gcdof2numbers(b, a%b)
}

func lcmof2numbers(a int, b int) int {
	return a / gcdof2numbers(a, b) * b
}

// マイナスの場合は考慮していない
func fuctorial(a int) int {
	if a == 1 || a == 0 {
		return 1
	} else {
		return fuctorial(a-1) * a
	}
}

func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}

// 素因数分解したmapを返す
func primeFuctorize(n int) map[int]int {
	pf := map[int]int{}
	for i := 2; i*i <= n; i++ {
		for n%i == 0 {
			pf[i]++
			n /= i
		}
	}
	if n != 1 {
		pf[n]++
	}
	return pf
}