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 Ruby (2.3.3) (judge verdict: Accepted)

n, k = gets.split.map(&:to_i)

nums = gets.split.map(&:to_i)

DIV = 10**9 + 7

$FACT_MUL = {}

def fact(num, i) 
    res = num
    mul = i * (i + 1) / 2
    # if $FACT_MUL.has_key?(i)
    #     mul = $FACT_MUL[i]
    # else
    #     1.upto(i) do |j|
    #         mul += j
    #         $FACT_MUL[j] = mul
    #     end
    # end
    res * [mul, 1].max
end

res = 0
nums.each_with_index do |outer_num, outer_idx|
    left_num = 0
    right_num = 0
    
    nums.each_with_index do |inner_num, inner_idx |
        if outer_num > inner_num
            if inner_idx < outer_idx
                left_num += 1
            else
                right_num += 1
            end
        end
    end
    res += fact(right_num, k)
    res += fact(left_num, k - 1)
end

puts res % DIV