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 "fmt"

func main() {
	var N, K, a, out int
	fmt.Scanf("%d %d", &N, &K)
	A := make([]int, N)
	for i := 0; i < N; i++ {
		fmt.Scanf("%d", &a)
		A[i] = a
	}
	out = 0
	var h, r int // A内の転倒数, A内で自分より小さい値の個数
	// 成分番号ごとにカウントする
	for i := 0; i < N; i++ {
		// h,rを計算
		h = 0
		r = 0
		for ii := 0; ii < N; ii++ {
			if A[ii] < A[i] {
				h++
				if ii > i {
					r++
				}
			}
		}
		//fmt.Printf("h:%d\n", h)
		//fmt.Printf("r:%d\n", r)
		out += h*K + r*(K-1)*K/2
		//fmt.Printf("h:%d\n", h)
		//fmt.Printf("r%d\n", r)
		//fmt.Printf("K%d\n", K)
		//fmt.Printf("out:%d\n", out)
	}
	//fmt.Println(A)
	// fmt.Printf("out:%d\n", out)
	fmt.Println(out % (1000000000 + 7))
}