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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

integer(16) N,K,i,j
integer(16),allocatable,dimension(:)::A,B,C
integer(16) cnt
integer(16) ans
read*,N,K
allocate(A(N),B(N),C(N))
read*,A
do i=1,N
  cnt=0
  do j=i+1,N
    if(A(i)>A(j))cnt=cnt+1
  end do
  B(i)=cnt
end do
do i=1,N
  cnt=0
  do j=1,N
    if(A(i)>A(j))cnt=cnt+1
  end do
  C(i)=cnt
end do
ans=0
do i=1,N
  ans=mod(ans+B(i)*K,10**9+7)
  ans=mod(ans+C(i)*((K-1)*K)/2,10**9+7)
end do
print"(i0)",ans
end