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.chomp.split(" ").map(&:to_i)
a = gets.chomp.split(" ").map(&:to_i)

num1 = Array.new(2000,0)
for i in 1..a.size do
    num1[a[-i]-1] += 1
end

num2 = Array.new(2000,0)

for i in 1..1999 do
    num2[i] = num2[i-1] + num1[i-1]
end

num1 = Array.new(2000,0)
cnt = 0
for i in 1..a.size do
    # print "--",a[-i],"--\n"
    tar = a[-i]-1
    a0 = 0
    d = num2[tar]
    for j in 1..a[-i]-1 do
        a0 += num1[j-1]
    end
    cnt += a0*k + d*k*(k-1)/2
    # print "a0 = ",a0,"\n"
    # print "d = ", d,"\n"
    num1[a[-i]-1] += 1
    # p cnt
end

# p ""
p cnt % (10**9 +7)