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"
	"time"
)

const devide int = 1000000007

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)

	sc.Scan()
	N, _ := strconv.Atoi(sc.Text())

	sc.Scan()
	K, _ := strconv.Atoi(sc.Text())

	A := make([]int, N)
	for i := 0; i < N; i++ {
		sc.Scan()
		A[i], _ = strconv.Atoi(sc.Text())
	}

	posts := 0
	alls := 0

	start := time.Now()
	for i := 0; i < N; i++ {
		for j := 0; j < i; j++ {
			if A[i] > A[j] {
				alls++
			}
		}

		for j := i + 1; j < N; j++ {
			if A[i] > A[j] {
				alls++
				posts++
			}
		}
	}

	result := posts
	pres := posts
	for i := 1; i < K; i++ {
		pres += alls
		result += pres
		result %= devide
	}
	fmt.Println(result)
	fmt.Println(time.Since(start))
}