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 Common Lisp (SBCL 1.1.14) (judge verdict: Accepted)

(let ((n (read))
      (k (read))
      (a (make-array 0 :element-type 'integer
                       :adjustable t
                       :fill-pointer t))
      (ans 0)
      kp kd)
  (dotimes (i n)
    (vector-push-extend (read) a))
  (setf kp (floor (* k (1+ k)) 2))
  (setf kd (floor (* k (1- k)) 2))
  (loop for i from 0 upto (1- n)
        do (let ((m (aref a i))
                 (cnt 0)
                 (mcnt 0))
             (loop for j from 0 upto (1- n)
                   do (cond ((and (< j i) (< (aref a j) m)) (incf mcnt))
                            ((< (aref a j) m) (incf cnt))))
             (incf ans (* cnt kp))
             (incf ans (* mcnt kd))
             (setf ans (rem ans 1000000007))))
  (princ ans))