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)

package main

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

func main() {
	var N, K int
	fmt.Scan(&N)
	fmt.Scan(&K)
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	var A []int
	for i := 0; i < N; i++ {
		sc.Scan()
		a, _ := strconv.Atoi(sc.Text())
		A = append(A, a)
	}

	numOfRightSmaller := make([]int64, N)
	numOfSmaller := make([]int64, N)
	for i := 0; i < N-1; i++ {
		for j := i + 1; j < N; j++ {
			if A[i] > A[j] {
				numOfRightSmaller[i]++
			}
		}
	}
	for i := 0; i < N; i++ {
		for j := 0; j < N; j++ {
			if A[i] > A[j] {
				numOfSmaller[i]++
			}
		}
	}

	var ans int64
	for i := 0; i < N; i++ {
		k := int64(K)
		tmp := (k * (k - 1) / 2) % (1E9 + 7)
		ans += numOfSmaller[i]*tmp + numOfRightSmaller[i]*k
		ans %= 1E9 + 7
	}
	fmt.Println(ans)
}